Category: Class 12

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Miscellaneous Exercise 6A Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Miscellaneous Exercise 6A

Question 1.
Find the vector equation of the line passing through the point having position vector \(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and parallel to \(6 \hat{i}-\hat{j}+\hat{k}\).
Solution:
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector
\(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and parallel to the vector \(6 \hat{i}-\hat{j}+\hat{k}\) is
\(\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+\lambda(\hat{6 \hat{\mathrm{i}}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\).

Question 2.
Find the vector equation of the line which passes through the point (3, 2, 1) and is parallel to the vector \(2 \hat{i}+2 \hat{j}-3 \hat{k}\).
Solution:
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector \(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and parallel to the vector
\(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \text { is } \overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})\)

Question 3.
Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and parallel to the line \(\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}\)
Solution:
The line \(\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}\) has direction ratios 3, 5, 6. The required line has direction ratios 3, 5, 6 as it is parallel to the given line.
It passes through the point (-2, 4, -5).
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are

Question 4.
Obtain the vector equation of the line \(\frac{x+5}{3}=\frac{y+4}{5}=\frac{z+5}{6}\).
Solution:
The cartesian equations of the line are \(\frac{x+5}{3}=\frac{y+4}{5}=\frac{z+5}{6}\).
This line is passing through the point A(-5, -4, -5) and having direction ratios 3, 5, 6.
Let \(\bar{a}\) be the position vector of the point A w.r.t. the origin and \(\bar{b}\) be the vector parallel to the line.
Then \(\bar{a}=-5 \hat{i}-4 \hat{j}-5 \hat{k}\) and \(\bar{b}=3 \hat{i}+5 \hat{j}+6 \hat{k}\).
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where λ is a scalar.
∴ the vector equation of the required line is \(\bar{r}=(-5 \hat{i}-4 \hat{j}-6 \hat{k})+\lambda(3 \hat{i}+5 \hat{j}+6 \hat{k})\)

Question 5.
Find the vector equation of the line which passes through the origin and the point (5, -2, 3).
Solution:
Let \(\bar{b}\) be the position vector of the point B(5, -2, 3).
Then \(\bar{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}\)
Origin has position vector \(\overline{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}\).
The vector equation the line passing through A(\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\) where λ is a scalar.
∴ the vector equation of the required line is \(\bar{r}=\overline{0}+\lambda(\bar{b}-\overline{0})=\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})\)

Question 6.
Find the Cartesian equations of the line which passes through points (3, -2, -5) and (3, -2, 6).
Solution:
Let A = (3, -2, -5), B = (3, -2, 6)
The direction ratios of the line AB are
3 – 3, -2 – (-2), 6 – (-5) i.e. 0, 0, 11.
The parametric equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
∴ the parametric equattions of the line passing through (3, -2, -5) and having direction ratios are 0, 0, 11 are
x = 3 + (0)λ, y = -2 + 0(λ), z = -5 + 11λ
i.e. x = 3, y = -2, z = 11λ – 5
∴ the cartesian equations of the line are
x = 3, y = -2, z = 11λ – 5, λ is a scalar.

Question 7.
Find the Cartesian equations of the line passing through A(3, 2, 1) and B(1, 3, 1).
Solution:
The direction ratios of the line AB are 3 – 1, 2 – 3, 1 – 1 i.e. 2, -1, 0.
The parametric equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
∴ the parametric equattions of the line passing through (3, 2, 1) and having direction ratios 2, -1, 0 are
x = 3 + 2λ, y = 2 – λ, z = 1 + 0(λ)
x – 3 = 2λ, y – 2 = -λ, z = 1
∴ \(\frac{x-3}{2}=\frac{y-2}{-1}\) = λ, z = 1
∴ the cartesian equations of the line are
\(\frac{x-3}{2}=\frac{y-2}{-1}\), z = 1.

Question 8.
Find the Cartesian equations of the line passing through the point A(1, 1, 2) and perpendicular to vectors \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}\).
Solution:
Let the required line have direction ratios p, q, r. ,
It is perpendicular to the vectors \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}\).
∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1.
∴ p + 2q + r = 0, 3p + 2q – r = 0

∴ the required line has direction ratios -1, 1, -1.
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
∴ the cartesian equations of the line passing through the point (1, 1, 2) and having direction ratios -1, 1, -1 are
\(\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-2}{-1}\)

Question 9.
Find the Cartesian equations of the line which passes through the point (2, 1, 3) and perpendicular
to lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}\).
Solution:
Let the required line have direction ratios p, q, r.
It is perpendicular to the vector \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\).
∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1.
∴ p + 2q + r = 0, 3 + 2q – r = 0

∴ the required line has direction ratios 2, -7, 4.
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x=x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
∴ the cartesian equation of the line passing through the point (2, -7, 4) and having directions ratios 2, -7, 4 are
\(\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-2}{4}\)

Question 10.
Find the vector equation of the line which passes through the origin and intersect the line x – 1 = y – 2 = z – 3 at right angle.
Solution:
The given line is \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}\) = λ … (Say)
∴ coordinates of any point on the line are
x = λ + 1, y = λ + 2, z = λ + 3
∴ position vector of any point on the line is
(λ + 1)\(\hat{i}\) + (λ + 2)\(\hat{j}\) + (λ + 3)\(\hat{k}\) … (1)
If \(\bar{b}\) is parallel to the given line whose direction ratios are 1, 1, 1, then \(\bar{b}=\hat{i}+\hat{j}+\hat{k}\).
Let the required line passing through O meet the given line at M.
∴ position vector of M
= \(\bar{m}\) = (λ + 1)\(\hat{i}\) + (λ + 2)\(\hat{j}\) + (λ + 3)\(\hat{k}\) … [By (1)]
The required line is perpendicular to given line

The vector equation of the line passing through A(\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\), λ is a scalar.
∴ the vector equation of the line passing through o(\(\bar{o}\)) and M(\(\bar{m}\)) is
\(\bar{r}=\overline{0}+\lambda(\bar{m}-\overline{0})=\lambda \bar{m}=\lambda(-\hat{i}+\hat{k})\) where λ is a scalar.
Hence, vector equation of the required line is \(\).

Question 11.
Find the value of λ so that lines \(\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angle.
Solution:
The equations of the given lines are

Question 12.
Find the acute angle between lines \(\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-3}{2}\) and \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{1}\).
Solution:

Question 13.
Find the acute angle between lines x = y, z = 0 and x = 0, z = 0.
Solution:
The equations x = y, z = 0 can be written as \(\frac{x}{1}=\frac{y}{1}\), z = 0
∴ the direction ratios of the line are 1, 1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis J are 0, 1, 0.
∴ its directiton ratios are 0, 1, 0.
Let \(\bar{a}\) and \(\bar{b}\) be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0.

If θ is the acute angle between the lines, then

Question 14.
Find the acute angle between lines x = -y, z = 0 and x = 0, z = 0.
Solution:
The equations x = -y, z = 0 can be written as \(\frac{x}{1}=\frac{y}{1}\), z = 0.
∴ the direction ratios of the line are 1, 1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis are 0, 1, 0.
∴ its direction ratios are 0, 1, 0.
Let \(\bar{a}\) and \(\bar{b}\) be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0

Question 15.
Find the co-ordinates of the foot of the perpendicular drawn from the point (0, 2, 3) to the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\).
Solution:
Let P = (0, 2, 3)
Let M be the foot of the perpendicular drawn from P to the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\) = λ ……(Say)
The coordinates of any point on the line are given by
x = 5λ – 3, y = 2λ + 1, z = 3λ – 4
Let M = (5λ – 3, 2λ + 1, 3λ – 4) …(1)
The direction ratios of PM are
5λ – 3 – 0, 2λ + 1 – 2, 3λ – 4 – 3 i.e. 5λ – 3, 2λ – 1, 3λ – 7
Since, PM is perpendicular to the line whose direcction ratios are 5, 2, 3,
5(5λ – 3) + 2(2λ – 1) + 3(3λ – 7) = 0
25λ – 15 + 4λ – 2 + 9λ – 21 =0
38λ – 38 = 0 ∴ λ = 1
Substituting λ = 1 in (1), we get.
M = (5 – 3, 2 + 1, 3 – 4) = (2, 3, -1).
Hence, the coordinates of the foot of perpendicular are (2, 3, – 1).

Question 16.
By computing the shortest distance determine whether following lines intersect each other.
(i) \(\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-\hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}+2 \hat{j}-3 \hat{k})+\mu(\hat{i}+\hat{j}-2 \hat{k})\)
Solution:
The shortest distance between the lines


Shortest distance between the lines is 0.
∴ the lines intersect each other.

(ii) \(\frac{x-5}{4}=\frac{y-7}{5}=\frac{z+3}{5}\) and x – 6 = y – 8 = z + 2.
Solution:
The shortest distance between the lines

∴ x₁ = 5, y₁ = 7, z₁ = 3, x₂ = 6, y₂ = 8, z₂ = 2,
l₁ = 4, m₁ = 5, n₁ = 1, l₂ = 1, m₂ = -2, n₂ = 1

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6)
= -16 – 36 – 64
= -116
and
(m₁n₂ – m₂n₁)² + (l₂n₁ – l₁n₂)² + (l₁m₂ – l₂m₁)²
= (-6 + 2)² + (1 – 7)² + (1 – 7)² + (-14 + 6)²
= 16 + 36 + 64
= 116
Hence, the required shortest distance between the given lines

or
Shortest distance between the lines is 0.
∴ the lines intersect each other.

Question 17.
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(\frac{x-2}{1}=\frac{y+m}{2}=\frac{z-2}{1}\) intersect each other then find m.
Solution:

Here, (x₁, y₁, z₁) ≡ (1, -1, 1),
(x₂, y₂, z₂) ≡ (2, -m, 2),
a₁ = 2, b₁ = 3, c₁ = 4,
a₂ = 1, b₂ = 2, c₂ = 1
Substituting these values in (1), we get

∴ 1(3 – 8) – (1 – m)(2 – 4) + 1 (4 – 3) = 0
∴ -5 + 2 – 2m + 1 = 0
∴ -2m = 2
∴ m = -1.

Question 18.
Find the vector and Cartesian equations of the line passing through the point (-1, -1, 2) and parallel to the line 2x – 2 = 3y + 1 = 6z – 2.
Solution:
Let \(\bar{a}\) be the position vector of the point A (-1, -1, 2) w.r.t. the origin.
Then \(\bar{a}=-\hat{i}-\hat{j}+2 \hat{k}\)
The equation of given line is
x – 2 = 3y + 1 = 6z – 2.

The direction ratios of this line are

Question 19.
Find the direction cosines of the line \(\bar{r}=\left(-2 \hat{i}+\frac{5}{2} \hat{j}-\hat{k}\right)+\lambda(2 \hat{i}+3 \hat{j})\).
Solution:

Question 20.
Find the Cartesian equation of the line passing through the origin which is perpendicular to x – 1 = y – 2 = z – 1 and intersects the \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\).
Solution:
Let the required line have direction ratios a, b, c
Since the line passes through the origin, its cartesian equations are
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) …(1)
This line is perpendicular to the line
x – 1 = y – 2 = z – 1 whose direction ratios are 1, 1, 1.

∴ 1(4b – 3c) + 1(4a – 2c) + 1(3a – 2b) = 0
∴ 4b – 3c + 4a – 2c + 3a – 2b = 0
∴ 7a + 2b – 5c = 0
From (2) and (3), we get

Question 21.
Write the vector equation of the line whose Cartesian equations are y = 2 and 4x – 3z + 5 = 0.
Solution:
4x – 3z + 5 = 0 can be written as

This line passes through the point A(0, 2, \(\frac{5}{3}\)) position vector is \(\bar{a}=2 \hat{j}+\frac{5}{3} \hat{k}\)
Also the line has direction ratio 3, 0, 4.
If \(\bar{b}\) is a vector parallel to the line, then \(\bar{b}=3 \hat{i}+4 \hat{k}\)
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where λ is \(\bar{a}\) scalar,
∴ the vector equation of the required line is
\(\bar{r}=\left(2 \hat{j}+\frac{5}{3} \hat{k}\right)+\lambda(3 \hat{i}+4 \hat{k})\).

Question 22.
Find the co-ordinates of points on the line \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{2}\) which are at the distance 3 unit from the base point A(1, 2, 3).
Solution:
The cartesian equations of the line are \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{2}\) = λ
The coordinates of any point on this line are given by
x = λ + 1, y = -2λ + 2, z = 2λ + 3
Let M(λ + 1, -2λ + 2, 2λ + 3) … (1)
be the point on the line whose distance from A(1, 2, 3) is 3 units.

When λ = 1, M = (1 + 1, -2 + 2, 2 + 3) … [By (1)]
i. e. M = (2, 0, 5)
When λ = -1, M = (1 – 1, 2 + 2, -2 + 3) … [By (1)]
i. e. M = (0, 4, 1)
Hence, the coordinates of the required points are (2, 0, 5) and (0, 4, 1).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.4

Question 1.
Find the angle between planes \(\bar{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})\) = 13 and \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 31 .
Solution:
The acute angle θ between the planes \(\bar{r} \cdot \bar{n}_{1}\) = d₁ and \(\bar{r} \cdot \bar{n}_{2}\) = d₂ is given by

Question 2.
Find the acute angle between the line \(\bar{r} \cdot(\hat{i}+2 \hat{j}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-6 \hat{k})\) and the plane \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 0
Solution:
The acute angle θ between the line \(\bar{r}=\bar{a}+\lambda \bar{b}\) and the plane \(\bar{r} \cdot \bar{n}\) = d is given by

Question 3.
Show that lines \(\bar{r}=(2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\) and \(\bar{r}=(2 \hat{i}+6 \hat{j}+3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})\) are coplanar. Find the equation of the plane determined by them.
Solution:

= 2(-1) + 6(2) + 3(-1)
= -2 + 12 – 3 = 7
∴ \(\bar{a}_{1} \cdot\left(\bar{b}_{1} \times \bar{b}_{2}\right)=\bar{a}_{2} \cdot\left(\bar{b}_{1} \times \bar{b}_{2}\right)\)
Hence, the given lines are coplanar.
The plane determined by these lines is given by
∴ \(\bar{r} \cdot\left(\overline{b_{1}} \times \overline{b_{2}}\right)=\overline{a_{1}} \cdot\left(\overline{b_{1}} \times \overline{b_{2}}\right)\)
i.e. \(\bar{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})\)
Hence, the given lines are coplanar and the equation of the plane determined by these lines is
\(\bar{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})\) = 7

Question 4.
Find the distance of the point \(4 \hat{i}-3 \hat{j}+\hat{k}\) from the plane \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-6 \hat{k})\) = 21 .
Solution:
The distance of the point A(\(\bar{a}\)) from the plane \(\bar{r} \cdot \bar{n}=p\) is given by d = \(\frac{|\bar{a} \cdot \bar{n}-p|}{|n|}\) …(1)
Here, \(\bar{a}=4 \hat{i}-3 \hat{j}+\hat{k}\), \(\bar{n}=2 \hat{i}+3 \hat{j}-6 \hat{k}\), p = 21
∴ \(\bar{a} \cdot \bar{n}\) = \((4 \hat{i}-3 \hat{j}+\hat{k}) \cdot(2 \hat{i}+3 \hat{j}-6 \hat{k})\)
= (4)(2) + (-3)(3) + (1)(-6)
= 8 – 9 – 6 = -7
Also, \(\sqrt{2^{2}+3^{2}+(-6)^{2}}=\sqrt{49}\) = 7
∴ from (1), the required distance
= \(\frac{|-7-21|}{7}\) = 4units

Question 5.
Find the distance of the point (1, 1, -1) from the plane 3x + 4y – 12z + 20 = 0.
Solution:
The distance of the point (x₁, y₁, z₁) from the plane ax + by + cz + d = 0 is \(\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)
∴ the distance of the point (1, 1, -1) from the plane 3x + 4y – 12z + 20 = 0 is \(\left|\frac{3(1)+4(1)-12(-1)+20}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|\)
= \(\left|\frac{3+4+12+20}{\sqrt{9+16+144}}\right|=\frac{39}{\sqrt{169}}\)
= \(\frac{39}{13}\) = 3units

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.3

Question 1.
Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector \(2 \hat{i}+\hat{j}-2 \hat{k}\).
Solution:
If \(\hat{n}\) is a unit vector along the normal and p is the length of the perpendicular from origin to the plane, then the vector equation of the plane is \(\bar{r} \cdot \hat{n}\) = p

Question 2.
Find the perpendicular distance of the origin from the plane 6x – 2y + 3z – 7 = 0.
Solution:
The equation of the plane is
6x – 2y + 3z – 7 = 0
∴ its vector equation is
\(\bar{r} \cdot(6 \hat{i}-2 \hat{j}+3 \hat{k})\) = 7 ….(1)
where \(\bar{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ \(\bar{n}=6 \hat{i}-2 \hat{j}+3 \hat{k}\) is normal to the plane
\(|\bar{n}|=\sqrt{6^{2}+(-2)^{2}+3^{2}}=\sqrt{49}\) = 7
Unit vector along \(\bar{n}\) is

Comparing with normal form of equation of the plane \(\bar{r} \cdot \hat{n}\) = p, it follows that length of perpendicular from origin is 1 unit.
Alternative Method:
The equation of the plane is 6x – 2y + 3z – 7 = 0 i.e. 6x – 2y + 3z = 7

This is the normal form of the equation of plane.
∴ perpendicular distance of the origin from the plane is p = 1 unit.

Question 3.
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 6y – 3z = 63 .
Solution:
The equation of the plane is 2x + 6y – 3z = 63. Dividing each term by \(\sqrt{2^{2}+6^{2}+(-3)^{2}}=\sqrt{49}\) = 7, we get
\(\frac{2}{7} x+\frac{6}{7} y-\frac{3}{7} z=\frac{63}{7}\) = 9
This is the normal form of the equation of plane.
∴ the direction cosines of the perpendicular drawn from the origin to the plane are
l = \(\frac{2}{7}\), m = \(\frac{6}{7}\), n = \(-\frac{3}{7}\)
and length of perpendicular from origin to the plane is p = 9.
∴ the coordinates of the foot of the perpendicular from the origin to the plane are (lp, mp, np) i.e. \(\left(\frac{18}{7}, \frac{54}{7},-\frac{27}{7}\right)\).

Question 4.
Reduce the equation \(\bar{r} \cdot(3 \hat{i}+4 \hat{j}+12 \hat{k})\) = 78 to normal form and hence find
(i) the length of the perpendicular from the origin to the plane
(ii) direction cosines of the normal.
Solution:
The normal form of equation of a plane is \(\bar{r} \cdot \hat{n}\) = p where \(\hat{n}\) is unit vector along the normal and p is the length of perpendicular drawn from origin to the plane.

This is the normal form of the equation of plane. Comparing with \(\bar{r} \cdot \hat{n}\) = p,
(i) the length of the perpendicular from the origin to plane is 6.
(ii) direction cosines of the normal are \(\frac{3}{13}, \frac{4}{13}, \frac{12}{13}\).

Question 5.
Find the vector equation of the plane passing through the point having position vector \(\hat{i}+\hat{j}+\hat{k}\) and perpendicular to the vector \(4 \hat{i}+5 \hat{j}+6 \hat{k}\).
Solution:
The vector equation of the plane passing through the point A (\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\)
Here, \(\bar{a}=\hat{i}+\hat{j}+\hat{k}\), \(\bar{n}=4 \hat{i}+5 \hat{j}+6 \hat{k}\)
∴ \(\bar{a} \cdot \bar{n}\) = \((\hat{i}+\hat{j}+\hat{k}) \cdot(4 \hat{i}+5 \hat{j}+6 \hat{k})\)
= (1)(4) + (1)(5) + (1)(6)
= 4 + 5 + 6 = 15
∴ the vector equation of the required plane is \(\bar{r} \cdot(4 \hat{i}+5 \hat{j}+6 \hat{k})\) = 15.

Question 6.
Find the Cartesian equation of the plane passing through A( -1, 2, 3), the direction ratios of whose normal are 0, 2, 5.
Solution:
The cartesian equation of the plane passing ; through (x₁, y₁, z₁), the direction ratios of whose normal are a, b, c, is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
∴ the cartesian equation of the required plane is
0(x +1) + 2(y – 2) + 5(z – 3) = 0
i.e. 0 + 2y – 4 + 5z – 15 = 0
i.e. 2y + 5z = 19.

Question 7.
Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the XY plane.
Solution:
The cartesian equation of the plane passing through (x₁, y₁, z₁), the direction ratios of whose normal are a, b, c, is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
The required plane is parallel to XY-plane.
∴ it is perpendicular to Z-axis i.e. Z-axis is normal to the plane. Z-axis has direction ratios 0, 0, 1.
The plane passes through (7, 8, 6).
∴ the cartesian equation of the required plane is
0(x – 7) + 0(y – 8) + 1 (z – 6) = 0
i.e. z = 6.

Question 8.
The foot of the perpendicular drawn from the origin to a plane is M(1, 0, 0). Find the vector equation of the plane.
Solution:
The vector equation of the plane passing ; through A(\(\bar{a}\)) and perpendicular to \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\).
M(1, 0, 0) is the foot of the perpendicular drawn from ; origin to the plane. Then the plane is passing through M : and is perpendicular to OM.
If \(\bar{m}\) is the position vector of M, then \(\bar{m}\) = \(\hat{i}\)
Normal to the plane is
\(\bar{n}\) = \(\overline{\mathrm{OM}}\) = \(\hat{i}\)
\(\bar{m} \cdot \bar{n}=\hat{i} \cdot \hat{i}\) = 1
∴ the vector equation of the required plane is \(\bar{r} \cdot \hat{i}\) = 1

Question 9.
Find the vector equation of the plane passing through the point A(-2, 7, 5) and parallel to vectors \(\hat{4}-\hat{j}+3 \hat{k}\) and \(\hat{i}+\hat{j}+\hat{k}\).
Solution:
The vector equation of the plane passing through the point A(\(\bar{a}\)) and parallel to the vectors \(\bar{b}\) and \(\bar{c}\) is
\(\bar{r} \cdot(\bar{b} \times \bar{c})=\bar{a} \cdot(\bar{b} \times \bar{c})\) ….(1)

Question 10.
Find the Cartesian equation of the plane \(\bar{r}=(5 \hat{i}-2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})\)
Solution:
The equation \(\) represents a plane passing through a point having position vector \(\bar{a}\) and parallel to vectors \(\bar{b}\) and \(\bar{c}\).


∴ 5x – 2y – 3z = 38.
This is the cartesian equation of the required plane.

Question 11.
Find the vector equation of the plane which makes intercepts 1, 1, 1 on the co-ordinates axes.
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)), B(\(\bar{b}\)). C(\(\bar{c}\)), where A, B, C are non-collinear is \(\bar{r} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) = \(\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) … (1)
The required plane makes intercepts 1, 1, 1 on the coordinate axes.
∴ it passes through the three non-collinear points A (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)

∴ from (1), the vector equation of the required plane is \(\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.2

Question 1.
Find the length of the perpendicular from (2, -3, 1) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+1}{-1}\)
Solution:
Let PM be the perpendicular drawn from the point P (2, -3, 1) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+1}{-1}\) = λ …(Say)
The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 3λ, z = -1 – λ
Let the coordinates of M be
(-1 + 2λ, 3 + 3λ, -1 – λ) … (1)
The direction ratios of PM are
-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1
i.e. 2λ – 3, 3λ + 6, -λ – 2
The direction ratios of the given line are 2, 3, -1.
Since PM is perpendicular to the given line, we get
2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = 0
∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0
∴ 14λ + 14 = 0 ∴ λ = -1.
Put λ = -1 in (1), the coordinats of M are
(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0,0).
∴ length of perpendicular from P to the given line

Alternative Method:
We know that the perpendicular distance from the point P\(|\bar{\alpha}|\) to the line \(\bar{r}=\bar{a}+\lambda \vec{b}\) is given by

Substituting these values in (1), we get
length of perpendicular from P to given line

Question 2.
Find the co-ordinates of the foot of the perpendicular drawn from the point \(2 \hat{i}-\hat{j}+5 \hat{k}\) to the line \(\bar{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})\). Also find the length of the perpendicular.
Solution:
Let M be the foot of perpendicular drawn from the point P (\(2 \hat{i}-\hat{j}+5 \hat{k}\)) on the line
\(\bar{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})\).
Let the position vector of the point M be

Then \(\overline{\mathrm{PM}}\) = Position vector of M – Position vector of P
= [(11 + 10λ)\(\hat{i}\) + (-2 – 4λ)\(\hat{j}\) + -8 – 11λ) \(\hat{k}\)] – (2\(\hat{i}\) – \(\hat{j}\) + 5\(\hat{k}\))
= (9 + 10λ)\(\hat{i}\) + (-1 – 4λ)\(\hat{j}\) + (-13 – 11λ)\(\hat{k}\)
Since PM is perpendicular to the given line which is parallel to \(\bar{b}=10 \hat{i}-4 \hat{j}-11 \hat{k}\),
\(\overline{\mathrm{PM}}\) ⊥^r\(\bar{b}\) ∴ \(\overline{\mathrm{PM}} \cdot \bar{b}\) = 0
∴ [(9 + 10λ)\(\hat{i}\) + ( – 1 – 4λ)\(\hat{j}\) + (-13 – 11λ)\(\hat{k}\)]-(10\(\hat{i}\) – 4\(\hat{j}\) – 11\(\hat{k}\)) = 0
∴ 10(9 +10λ) – 4( -1 – 4λ) – 11( -13 – 11λ) = 0
∴ 90 + 100λ + 4 + 16λ + 143 +121λ = 0
∴ 237λ + 237 = 0
∴ λ = -1
Putting this value of λ, we get the position vector of M as \(\hat{i}+2 \hat{j}+3 \hat{k}\).
∴ coordinates of the foot of perpendicular M are (1, 2, 3).

Hence, the coordinates of the foot of perpendicular are (1,2, 3) and length of perpendicular = \(\sqrt {14}\) units.

Question 3.
Find the shortest distance between the lines \(\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})\) and \(\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+4 \hat{j}-5 \hat{k})\)
Solution:
We know that the shortest distance between the skew lines \(\bar{r}=\overline{a_{1}}+\lambda \overline{b_{1}}\) and \(\bar{r}=\overline{a_{2}}+\mu \overline{b_{2}}\) is given by

Question 4.
Find the shortest distance between the lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Solution:
The shortest distance between the lines

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6)
= -16 – 36 – 64 = -116
and (m₁n₂ – m₂n₁)² + (l₂n₁ – l₁n₂)² + (l₁m₂ – l₂m₁)²
= (-6 + 2)² + (1 – 7)² + (-14 + 6)²
= 16 + 36 + 64 = 116
Hence, the required shortest distance between the given lines = \(\left|\frac{-116}{\sqrt{116}}\right|\) = \(\sqrt{116}\) = \(2 \sqrt{29}\) units

Question 5.
Find the perpendicular distance of the point (1, 0, 0) from the line \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) Also find the co-ordinates of the foot of the perpendicular.
Solution:
Let PM be the perpendicular drawn from the point (1, 0, 0) to the line \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) = λ …(Say)
The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 2λ, z = 8 – λ
Let the coordinates of M be
(-1 + 2λ, 3 + 3λ, -1 – λ) …..(1)
The direction ratios of PM are
-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1
i.e. 2λ – 3, 3λ = 6, -λ – 2
The direction ratios of the given line are 2, 3, 8.
Since PM is perpendicular to the given line, we get
2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = O
∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0
∴ 14λ + 14 = 0
∴ λ = -1
Put λ in (1), the coordinates of M are
(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0, 0).
∴ length of perpendicular from P to the given line
= PM
= \(\sqrt{(-3-2)^{2}+(0+3)^{2}+(0-1)^{2}}\)
= \(\sqrt{(25 + 9 + 1)}\)
= \(\sqrt{35}\)units.
Alternative Method :
We know that the perpendicular distance from the point

Substitutng tese values in (1), w get
length of perpendicular from P to given line
= PM

Question 6.
A(1, 0, 4), B(0, -11, 13), C(2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.
Solution:
Equation of the line passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) is

AD is the perpendicular from the point A (1, 0, 4) to the line BC.
The coordinates of any point on the line BC are given by x = 2λ, y = -11 + 8λ, z = 13 – 12λ
Let the coordinates of D be (2λ, -11 + 8λ, 13 – 12λ) … (1)
∴ the direction ratios of AD are
2λ – 1, -1λ + 8λ – 0, 13 – 12λ – 4 i.e.
2λ – 1, -11 + 8λ, 9 – 12λ
The direction ratios of the line BC are 2, 8, -12.
Since AD is perpendicular to BC, we get
2(2λ – 1) + 8(-11 + 8λ) – 12(9 – 12λ) = 0
∴ 42λ – 2 – 88 + 64λ – 108 + 144λ = 0
∴ 212λ – 198 = 0

Question 7.
By computing the shortest distance, determine whether following lines intersect each other.
(i) \(\bar{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}+\hat{j}-\hat{k})\)
Solution:
The shortest distance between the lines


Hence, the given lines do not intersect.

(ii) \(\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}\) and \(\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}\)
Solution:
The shortest distance between the lines

∴ x₁ = -1, y₁ = -1, z₁ = -1, x₂ = 3, y₂ = 5, z₂ = 7,
l₁ = 7, m₁ = -6, n₁ = 1, l₂ = 1, m₂ = -2, n₂ = 1
\(\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
l_{1} & m_{1} & n_{1} \\
l_{2} & m_{2} & n_{2}
\end{array}\right|\) = \(\left|\begin{array}{ccc}
4 & 6 & 8 \\
4 & -5 & -5 \\
7 & 1 & 3
\end{array}\right|\)
= 4(- 6 + 2) – 6(7 – 1) + 8(-14 + 6)
= -16 – 36 – 64
= -116
and
(m₁n₂ – m₂n₁)² + (l₂n₁ – l₁n₂)² + (l₁m₂ – l₂m₁)²
= (-6 + 2)² + (1 – 7)² + (-14 + 6)²
= 16 + 36 + 64
= 116
Hence, the required shortest distance between the given lines
= \(\left|\frac{-116}{\sqrt{116}}\right|\)
= \(\sqrt{116}\)
=\(2 \sqrt{29}\) units
or
The shortest distance between the lines
= \(\frac{282}{\sqrt{3830}}\)units
Hence, the gives lines do not intersect.

Question 8.
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect each other then find k.
Solution:
The lines

∴ x₁ = 1, y₁ = -1, z₁ = 1, x₂ = 3, y₂ = k, z₂ = 0,
l₁ = 2, m₁ = 3, n₁ = 4, l₂ = 1, m₂ = 2, n₂ = 1.
Since these lines intersect, we get
\(\left|\begin{array}{ccc}
2 & k+1 & -1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|\) = 0
∴ 2 (3 – 8) – (k + 1)(2 – 4) – 1 (4 – 3) = 0
∴ -10 + 2(k + 1) – 1 = 0
∴ 2(k + 1) = 11
∴ k + 1 = \(\frac{11}{2}\)
∴ k = \(\frac{9}{2}\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.1

Question 1.
Find the vector equation of the line passing through the point having position vector \(-2 \hat{i}+\hat{j}+\hat{k}\) and parallel to vector \(4 \hat{i}-\hat{j}+2 \hat{k}\).
Solution:
The vector equation of the line passing through A (\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\bar{r}\) = \(\bar{a}\) + λ\(\bar{b}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector \(-2 \hat{i}+\hat{j}+\hat{k}\) and parallel to the vector \(4 \hat{i}-\hat{j}+2 \hat{k}\) is
\(\bar{r}=(-2 \hat{i}+\hat{j}+\hat{k})+\lambda(4 \hat{i}-\hat{j}+2 \hat{k})\).

Question 2.
Find the vector equation of the line passing through points having position vectors \(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and \(6 \hat{i}-\hat{j}+\hat{k}\).
Solution:
The vector equation of the line passing through the A (\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\), λ is a scalar
∴ the vector equation of the line passing through the points having position vectors \(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and \(6 \hat{i}-\hat{j}+\hat{k}\) is

Question 3.
Find the vector equation of line passing through the point having position vector \(5 \hat{i}+4 \hat{j}+3 \hat{k}\) and having direction ratios -3, 4, 2.
Solution:
Let A be the point whose position vector is \(\bar{a}=5 \hat{i}+4 \hat{j}+3 \hat{k}\).
Let \(\bar{b}\) be the vector parallel to the line having direction ratios -3, 4, 2
Then, \(\bar{b}\) = \(-3 \hat{i}+4 \hat{j}+2 \hat{k}\)
The vector equation of the line passing through A (\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}=\bar{a}+\lambda \bar{b}\), where λ is a scalar.
∴ the required vector equation of the line is
\(\bar{r}=5 \hat{i}+4 \hat{j}+3 \hat{k}+\lambda(-3 \hat{i}+4 \hat{j}+2 \hat{k})\).

Question 4.
Find the vector equation of the line passing through the point having position vector \(\hat{i}+2 \hat{j}+3 \hat{k}\) and perpendicular to vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(2 \hat{i}-\hat{j}+\hat{k}\).
Solution:

Since the line is perpendicular to the vector \(\bar{b}\) and \(\bar{c}\), it is parallel to \(\bar{b} \times \bar{c}\). The vector equation of the line passing through A (\(\bar{a}\)) and parallel to \(\bar{b} \times \bar{c}\) is
\(\bar{r}=\bar{a}+\lambda(\bar{b} \times \bar{c})\), where λ is a scalar.
Here, \(\bar{a}\) = \(\hat{i}+2 \hat{j}+3 \hat{k}\)
Hence, the vector equation of the required line is
\(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})\)

Question 5.
Find the vector equation of the line passing through the point having position vector \(-\hat{i}-\hat{j}+2 \hat{k}\) and parallel to the line \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})\).
Solution:
Let A be point having position vector \(\bar{a}\) = \(-\hat{i}-\hat{j}+2 \hat{k}\)
The required line is parallel to the line

The vector equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}\) = \(\bar{a}\) + λ\(\bar{b}\) where λ is a scalar.
∴ the required vector equation of the line is
\(\overline{\mathrm{r}}=(-\hat{i}-\hat{j}+2 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})\).

Question 6.
Find the Cartesian equations of the line passing through A(-1, 2, 1) and having direction ratios 2, 3, 1.
Solution:
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are

∴ the cartesian equations of the line passing through the point (-1, 2, 1) and having direction ratios 2, 3, 1 are

Question 7.
Find the Cartesian equations of the line passing through A(2, 2, 1) and B(1, 3, 0).
Solution:
The cartesian equations of the line passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are

Here, (x₁, y₁, z₁) = (2, 2, 1) and (x₂, y₂, z₂) = (1, 3, 0)
∴ the required cartesian equations are

Question 8.
A(-2, 3, 4), B(1, 1, 2) and C(4, -1, 0) are three points. Find the Cartesian equations of the line AB and show that points A, B, C are collinear.
Solution:
We find the cartesian equations of the line AB. The cartesian equations of the line passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are
\(\frac{x-x_{1}}{x_{2}-x_{1}}\) = \(\frac{y-y_{1}}{y_{2}-y_{1}}\) = \(\frac{z-z_{1}}{z_{2}-z_{1}}\)
Here, (x₁, y₁, z₁) = (-2, 3, 4) and (x₂, y₂, z₂) = (4, -1, 0)
∴ the required cartesian equations of the line AB are


∴ coordinates of C satisfy the equations of the line AB.
∴ C lies on the line passing through A and B.
Hence, A, B, C are collinear.

Question 9.
Show that lines \(\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}\) and \(\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}\) intersect each other. Find the co-ordinates of their point of intersection.
Solution:
The equations of the lines are

From (1), x = -1 -10λ, y = -3 – 2, z = 4 + λ
∴ the coordinates of any point on the line (1) are
(-1 – 10λ, – 3 – λ, 4 + λ)
From (2), x = -10 – u, y = -1 – 3u, z = 1 + 4u
∴ the coordinates of any point on the line (2) are
(-10 – u, -1 – 3u, 1 + 4u)
Lines (1) and (2) intersect, if
(- 1 – 10λ, – 3 – λ, 4 + 2) = (- 10 – u, -1 – 3u, 1 + 4u)
∴ the equations -1 – 10λ = -10 – u, -3 – 2= – 1 – 3u
and 4 + λ = 1 + 4u are simultaneously true.
Solving the first two equations, we get, λ = 1 and u = 1. These values of λ and u satisfy the third equation also.
∴ the lines intersect.
Putting λ = 1 in (-1 – 10λ, -3 – 2, 4 + 2) or u = 1 in (-10 – u, -1 – 3u, 1 + 4u), we get
the point of intersection (-11, -4, 5).

Question 10.
A line passes through (3, -1, 2) and is perpendicular to lines \(\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\mu(\hat{i}-2 \hat{j}+2 \hat{k})\). Find its equation.
Solution:

The vector perpendicular to the vectors \(\bar{b}\) and \(\bar{c}\) is given by

Since the required line is perpendicular to the given lines, it is perpendicular to both \(\bar{b}\) and \(\bar{c}\).
∴ it is parallel to \(\bar{b} \times \bar{c}\)
The equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b} \times \bar{c}\) is
\(\bar{r}=\bar{a}+\lambda(\bar{b} \times \bar{c})\), where λ is a scalar.
Here, \(\bar{a}\) = \(3 \hat{i}-\hat{j}+2 \hat{k}\)
∴ the equation of the required line is

Question 11.
Show that the line \(\frac{x-2}{1}=\frac{y-4}{2}=\frac{z+4}{-2}\) passes through the origin.
Solution:
The equation of the line is
\(\frac{x-2}{1}=\frac{y-4}{2}=\frac{z+4}{-2}\)
The coordinates of the origin O are (0, 0, 0)

∴ coordinates of the origin O satisfy the equation of the line.
Hence, the line passes through the origin.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Miscellaneous Exercise 5

I) Select the correct option from the given alternatives :
Question 1.
If |\(\bar{a}\)| = 2, |\(\bar{b}\)| = 3 |\(\bar{c}\)| = 4 then [\(\bar{a}\) + \(\bar{b}\) \(\bar{b}\) + \(\bar{c}\) \(\bar{c}\) – \(\bar{a}\)] is equal to
(A) 24
(B) -24
(C) 0
(D) 48
Solution:
(C) 0

Question 2.
If |\(\bar{a}\)| = 3, |\(\bar{b}\)| = 4, then the value of λ for which \(\bar{a}\) + λ\(\bar{b}\) is perpendicular to \(\bar{a}\) – λ\(\bar{b}\), is

Solution:
(b) \(\frac{3}{4}\)

Question 3.
If sum of two unit vectors is itself a unit vector, then the magnitude of their difference is
(A) \(\sqrt {2}\)
(B) \(\sqrt {3}\)
(C) 1
(D) 2
Solution:
(B) \(\sqrt {3}\)

Question 4.
If |\(\bar{a}\)| = 3, |\(\bar{b}\)| = 5, |\(\bar{c}\)| = 7 and \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0, then the angle between \(\bar{a}\) and \(\bar{b}\) is

Solution:
(b) \(\frac{\pi}{3}\)

Question 5.
The volume of tetrahedron whose vertices are (1, -6, 10), (-1, -3, 7), (5, -1, λ) and (7, -4, 7) is 11 cu. units then the value of λ is
(A) 7
(B) \(\frac{\pi}{3}\)
(C) 1
(D) 5
Solution:
(A) 7

Question 6.
If α, β, γ are direction angles of a line and α = 60º, β = 45º, the γ =
(A) 30º or 90º
(B) 45º or 60º
(C) 90º or 30º
(D) 60º or 120º
Solution:
(D) 60º or 120º

Question 7.
The distance of the point (3, 4, 5) from Y- axis is
(A) 3
(B) 5
(C) \(\sqrt {34}\)
(D) \(\sqrt {41}\)
Solution:
(C) \(\sqrt {34}\)

Question 8.
The line joining the points (-2, 1, -8) and (a, b, c) is parallel to the line whose direction ratios are 6, 2, 3. The value of a, b, c are
(A) 4, 3, -5
(B) 1, 2, \(\frac{-13}{2}\)
(C) 10, 5, -2
(D) 3, 5, 11
Solution:
(A) 4, 3, -5

Question 9.
If cos α, cos β, cos γ are the direction cosines of a line then the value of sin² α + sin²β + sin²γ is
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
(B) 2

Question 10.
If l, m, n are direction cosines of a line then \(\hat{l}+m \hat{j}+n \hat{k}\) is
(A) null vector
(B) the unit vector along the line
(C) any vector along the line
(D) a vector perpendicular to the line
Solution:
(B) the unit vector along the line

Question 11.
If |\(\bar{a}\)| = 3 and –1 ≤ k ≤ 2, then |k\(\bar{a}\)| lies in the interval
(A) [0, 6]
(B) [-3, 6]
(C) [3, 6]
(D) [1, 2]
Solution:
(A) [0, 6]

Question 12.
Let α, β, γ be distinct real numbers. The points with position vectors \(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\), \(\beta \hat{i}+\gamma \hat{j}+\alpha \hat{k}\), \(\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}\)
(A) are collinear
(B) form an equilateral triangle
(C) form a scalene triangle
(D) form a right angled triangle
Solution:
(B) form an equilateral triangle

Question 13.
Let \(\bar{p}\) and \(\bar{q}\) be the position vectors of P and Q respectively, with respect to O and |\(\bar{p}\)| = p, |\(\bar{q}\)| = q. The points R and S divide PQ internally and externally in the ratio 2 : 3 respectively. If OR and OS are perpendicular then.
(A) 9p² = 4q²
(B) 4p² = 9q²
(C) 9p = 4q
(D) 4p = 9q
Solution:
(A) 9p² = 4q²

Question 14.
The 2 vectors \(\hat{j}+\hat{k}\) and \(3 \hat{i}-\hat{j}+4 \hat{k}\) represents the two sides AB and AC, respectively of a ∆ABC. The length of the median through A is
(A) \(\frac{\sqrt{34}}{2}\)
(B) \(\frac{\sqrt{48}}{2}\)
(C) \(\sqrt {18}\)
(D) None of these
Solution:
(A) \(\frac{\sqrt{34}}{2}\)

Question 15.
If \(\bar{a}\) and \(\bar{b}\) are unit vectors, then what is the angle between \(\bar{a}\) and \(\bar{b}\) for \(\sqrt{3} \bar{a}\) – \(\bar{b}\) to be a unit vector ?
(A) 30º
(B) 45º
(C) 60º
(D) 90º
Solution:
(A) 30º

Question 16.
If θ be the angle between any two vectors \(\bar{a}\) and \(\bar{b}\), then \(|\vec{a} \cdot \vec{b}|\) = \(|\vec{a} \times \vec{b}|\), when θ is equal to
(A) 0
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{2}\)
(D) π
Solution:
(B) \(\frac{\pi}{4}\)

Question 17.
The value of \(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})\)
(A) 0
(B) -1
(C) 1
(D) 3
Solution:
(C) 1

Question 18.
Let a, b, c be distinct non-negative numbers. If the vectors \(\mathrm{a} \hat{i}+\mathrm{a} \hat{j}+\mathrm{c} \hat{k}\), \(\hat{i}+\hat{k}\) and \(\mathbf{c} \hat{i}+\mathrm{c} \hat{j}+\mathrm{b} \hat{k}\) lie in a plane, then c is
(A) The arithmetic mean of a and b
(B) The geometric mean of a and b
(C) The harmonic man of a and b
(D) 0
Solution:
(B) The geometric mean of a and b

Question 19.
Let \(\bar{a}\) = \(\hat{i} \hat{j}\), \(\bar{b}\) = \(\hat{j} \hat{k}\), \(\bar{c}\) = \(\hat{k} \hat{i}\). If \(\bar{d}\) is a unit vector such that \(\bar{a} . \bar{d}=0=\left[\begin{array}{lll}
\bar{b} & \bar{c} & \bar{d}
\end{array}\right]\), then \(\bar{d}\) equals.

Solution:
(a) \(\pm \frac{\hat{i}+\hat{j}-2 \hat{k}}{\sqrt{6}}\)

Question 20.
If \(\bar{a}\) \(\bar{b}\) \(\bar{c}\) are non coplanar unit vectors such that \(\bar{a} \times(\bar{b} \times \bar{c})\) =\(\frac{(\bar{b}+\bar{c})}{\sqrt{2}}\) then the angle between \(\bar{a}\) and \(\bar{b}\) is
(A) \(\frac{3 \pi}{4}\)
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{2}\)
(D) π
Solution:
(A) \(\frac{3 \pi}{4}\)

II Answer the following :
1) ABCD is a trapezium with AB parallel to DC and DC = 3AB. M is the mid-point of DC,
\(\overline{A B}\) = \(\bar{p}\) and \(\overline{B C}\) = \(\bar{q}\). Find in terms of \(\bar{p}\) and \(\bar{q}\).
(i) \(\overline{A M}\)
Solution:

(ii) \(\overline{B D}\)
Solution:

(iii) \(\overline{M B}\)
Solution:

(iv) \(\overline{D A}\)
Solution:

Question 2.
The points A, B and C have position vectors \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) respectively. The point P is midpoint of AB. Find in terms of \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) the vector \(\overline{P C}\)
Solution:
P is the mid-point of AB.
∴ \(\bar{p}\) = \(=\frac{\bar{a}+\bar{b}}{2}\), where \(\bar{p}\) is the position vector of P.

Question 3.
In a pentagon ABCDE
Show that \(\overline{A B}\) + \(\overline{A E}\) + \(\overline{B C}\) + \(\overline{D C}\) + \(\overline{E D}\) = 2\(\overline{A C}\)
Solution:

Question 4.
If in parallelogram ABCD, diagonal vectors are \(\overline{A C}\) = \(2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\overline{B D}\) = \(-6 \hat{i}+7 \hat{j}-2 \hat{k}\), then find the adjacent side vectors \(\overline{A B}\) and \(\overline{A D}\)
Solution:
ABCD is a parallelogram

Question 5.
If two sides of a triangle are \(\hat{i}+2 \hat{j}\) and \(\hat{i}+\hat{k}\), then find the length of the third side.
Solution:

Let ABC be a triangle with \(\overline{A B}\) = \(\hat{i}+2 \hat{j}\), \(\overline{B C}\) = \(\hat{i}+\hat{k}\).
By triangle law of vectors

Hence, the length of third side is 3 units.

Question 6.
If |\(\bar{a}\)| = |\(\bar{b}\) | = 1 \(\bar{a}\).\(\bar{b}\) = 0 and \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0 then find |\(\bar{c}\)|
Solution:
\(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0
∴ –\(\bar{c}\) = \(\bar{a}\) + \(\bar{b}\)
Taking dot product of both sides with itself, we get

Question 7.
Find the lengths of the sides of the triangle and also determine the type of a triangle.
(i) A(2, -1, 0), B(4, 1, 1,), C(4, -5, 4)
Solution:
The position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) of the points A, B, C are


∴ ∆ ABC is right angled at A.

(ii) L(3, -2, -3), M(7, 0, 1), N (1, 2, 1)
Solution:
The position vectors bar \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) of the points L M, N are

l(LM) = 6, l(MN) = 2\(\sqrt {10}\) , l(NL) = 6
∆LMN is sosceles

Question 8.
Find the component form of if a if
(i) It lies in YZ plane and makes 60º with positive Y-axis and |\(\bar{a}\)| = 4
Solution:
Let α, β, γ be the direction angles of \(\bar{a}\)
Since \(\bar{a}\) lies in YZ-plane, it is perpendicular to X-axis
∴ α = 90°
It is given that β= 60°
∵ cos²α + cos²β + cos²γ = 1
∴ cos²90° + cos²60° + cos²γ = 1
∴ 0 + \(\left(\frac{1}{2}\right)^{2}\) + cos²γ = 1
∴ cos²γ = 1 – \(\frac{1}{4}=\frac{3}{4}\)
∴ cos γ = \(\pm \frac{\sqrt{3}}{2}\)
Unit vector along a is given by

(ii) It lies in XZ plane and makes 45º with positive Z-axis and |\(\bar{a}\)| = 10
Solution:

Question 9.
Two sides of a parallelogram are \(3 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(-2 \hat{j}+7 \hat{k}\). Find the unit vectors parallel to the diagonals.
Solution:

Question 10.
If D, E, F are the mid-points of the sides BC, CA, AB of a triangle ABC , prove that \(\overline{A D}\) + \(\overline{B E}\) + \(\overline{C F}\) = 0

Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\), \(\bar{e}\), \(\bar{f}\) be the position vectors of the points A, B, C, D, E, F respectively.
Since D, E, F are the midpoints of BC, CA, AB respec-tively, by the midpoint formula

Question 11.
Find the unit vectors that are parallel to the tangent line to the parabola y = x² at the point (2, 4)
Solution:
Differentiating y = x² w.r.t. x, we get \(\) = 2x
Slope of tangent at P(2, 4) = \(\left(\frac{d y}{d x}\right)_{\text {at } \mathrm{P}(2, 4)}\) = 2 × 2 = 4
∴ the equation of tangent at P is
y – 4 = 4(-2)
∴ y = 4x – 4
∴ y = 4x is equation of line parallel to the tangent at P and passing through the origin O.
4x = y, z = 0 ∴ \(\frac{x}{1}=\frac{y}{4}\), z = 0
∴ the direction ratios of this line are 1, 4, 0
∴ its direction cosines are

Question 12.
Express the vector \(\hat{i}+4 \hat{j}-4 \hat{k}\) as a linear combination of the vectors \(2 \hat{i}-\hat{j}+3 \hat{k}\), \(\hat{i}-2 \hat{j}+4 \hat{k}\) and \(-\hat{i}+3 \hat{j}-5 \hat{k}\)
Solution:

By equality of vectors,
2x + 2y – z = 1
-x – 2y + 3z = 4
3x + 4y – 5z = -4
We have to solve these equations by using Cramer’s Rule.
D = \(\left|\begin{array}{ccc}
2 & 2 & -1 \\
-1 & -2 & 3 \\
3 & 4 & -5
\end{array}\right|\)
= 2(10 – 12) – 2(5 – 9) – 1(-4 + 6)
= -4 + 8 – 2
= 2 ≠ 0
D_x = \(\left|\begin{array}{ccc}
1 & 2 & -1 \\
4 & -2 & 3 \\
-4 & 4 & -5
\end{array}\right|\)
= 1(10 – 12) – 2(-20 + 12) – 1 (16 – 8)
= -2 + 16 – 8
= 6
D_y = \(\left|\begin{array}{ccc}
2 & 1 & -1 \\
-1 & 4 & 3 \\
3 & -4 & -5
\end{array}\right|\)
= 2(-20 + 12) – 1(5 – 9) – 1(4 – 12)
= -16 – 4 – 8
= -28
D_z = \(\left|\begin{array}{ccc}
2 & 2 & 1 \\
-1 & -2 & 4 \\
3 & 4 & -4
\end{array}\right|\)
= 2(8 – 16) – 2(4 – 12) + 1(-4 + 6)
= -16 – 16 + 2
= -30

Question 13.
If \(\overline{O A}\) = \(\bar{a}\) and \(\overline{O B}\) = \(\bar{b}\) then show that the vector along the angle bisector of angle AOB is
given by \(\bar{d}\) = λ\(\left(\frac{\bar{a}}{|\bar{b}|}+\frac{\bar{b}}{|\bar{b}|}\right)\)
Question is modified
If \(\overline{O A}\) = \(\bar{a}\) and \(\overline{O B}\) = \(\bar{b}\) then show that the vector along the angle bisector of ∠AOB is
given by \(\bar{d}\) = λ\(\left(\frac{\bar{a}}{|\bar{a}|}+\frac{\bar{b}}{|\bar{b}|}\right)\)
Solution:

Choose any point P on the angle bisector of ∠AOB. Draw PM parallel to OB.
∴ ∠OPM = ∠POM
= ∠POB
Hence, OM = MP
∴ OM and MP is the same scalar multiple of unit vectors \(\hat{a}\) and \(\hat{b}\) along these directions,

Question 14.
The position vectors f three consecutive vertices of a parallelogram are \(\hat{i}+\hat{j}+\hat{k}\), \(\hat{i}+3 \hat{j}+5 \hat{k}\) and \(7 \hat{i}+9 \hat{j}+11 \hat{k}\) Find the position vector of the fourth vertex.
Solution:
Let ABCD be a parallelogram.
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\) be the position vectors of the vertices
A, B, C, D of the parallelogram,

Hence, the position vector of the fourth vertex is 7(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)).

Question 15.
A point P with position vector \(\frac{-14 \hat{i}+39 \hat{j}+28 \hat{k}}{5}\) divides the line joining A(-1, 6, 5) and B in the ratio 3 : 2 then find the point B.
Solution:
Let A, B and P have position vectors \(\bar{a}\), \(\bar{b}\) and \(\bar{p}\) respectively.

∴ coordinates of B are (-4, 9, 6).

Question 16.
Prove that the sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.
Solution:
Let \(\overrightarrow{\mathrm{a}}\), \(\overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{c}}\) are the position vectors of the vertices A, B and C respectively.
Then we know that the position vector of the centroid O of the triangle is \(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\)
Therefore sum of the three vectors \(\overrightarrow{\mathrm{OA}}\), \(\overrightarrow{\mathrm{OB}}\) and \(\overrightarrow{\mathrm{OC}}\), is

Hence, Sum os the three vectors determined by the medians of a triangle directed from the vertices is zero.

Question 17.
ABCD is a parallelogram E, F are the mid points of BC and CD respectively. AE, AF meet the diagonal BD at Q and P respectively. Show that P and Q trisect DB.
Solution:


LHS is the position vector of the point on AE and RHS is the position vector of the point on DB. But AE and DB meet at Q.

LHS is the position vector of the point on AF and RHS is the position vector of the point on DB.
But AF and DB meet at P.
∴ \(\bar{p}=\frac{\bar{b}+2 \bar{d}}{1+2}\)
∴ P divides DB in the ratio 1 : 2 … (5)
From (4) and (5), if follows that P and Q trisect DB.

Question 18.
If aBC is a triangle whose orthocenter is P and the circumcenter is Q, then prove that \(\overline{P A}\) + \(\overline{P C}\) + \(\overline{P B}\) = 2 \(\overline{P Q}\)
Solution:
Let G be the centroid of the ∆ ABC.
Let A, B, C, G, Q have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{g}\), \(\bar{q}\) w.r.t. P. We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2.

Question 19.
If P is orthocenter, Q is circumcenter and G is centroid of a triangle ABC, then prove that \(\overline{Q P}\) = 3\(\overline{Q G}\)
Solution:
Let \(\bar{p}\) and \(\bar{g}\) be the position vectors of P and G w.r.t. the circumcentre Q.
i.e. \(\overline{\mathrm{QP}}\) = p and \(\overline{\mathrm{QG}}\) = g.
We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2
∴ by section formula for internal division,

Question 20.
In a triangle OAB, E is the midpoint of BO and D is a point on AB such that AD: DB = 2:1. If OD and AE intersect at P, determine the ratio OP:PD using vector methods.
Solution:

Let A, B, D, E, P have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{d}\), \(\bar{e}\), \(\bar{p}\) respectively w.r.t. O.
∵ AD : DB = 2 : 1.
∴ D divides AB internally in the ratio 2 : 1.
Using section formula for internal division, we get

LHS is the position vector of the point which divides OD internally in the ratio 3 : 2.
RHS is the position vector of the point which divides AE internally in the ratio 4 : 1.
But OD and AE intersect at P
∴ P divides OD internally in the ratio 3 : 2.
Hence, OP : PD = 3 : 2.

Question 21.
Dot-product of a vector with vectors \(3 \hat{i}-5 \hat{k}, 2 \hat{i}+7 \hat{j}\) and \(\hat{i}+\hat{j}+\hat{k}\) are respectively -1, 6 and 5. Find the vector.
Solution:

∴ 3x – 5z= -1 … (1)
∴ 2x + 7y = 6 … (2)
∴ x + y + z = 5 … (3)
From (3), z = 5 – x – y
Substituting this value of z in (1), we get
∴ 3x – 5(5 – x – y)= -1
∴ 8x + 5y = 24 … (4)
Multiplying (2) by 4 and subtracting from (4), we get
8x + 5y – 4(2x + 7y) = 24 – 6 × 4
∴ -23y = 0 ∴ y = 0
Substituting y = 0 in (2), we get
∴ 2x = 6 ∴ x = 3
Substituting x = 3 in (1), we get
∴ 3(3) – 5z = -1
∴ 5z = -10 ∴ z = 2
∴ \(\bar{r}=3 \hat{i}+0 \cdot \hat{j}+2 \hat{k}=3 \hat{i}+2 \hat{k}\)
Hence, the required vector is \(3 \hat{i}+2 \hat{k}\)

Question 22.
If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are unit vectors such that \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0, then find the value of \(\bar{a}\).\(\bar{b}\) + \(\bar{b}\).\(\bar{c}\) + \(\bar{c}\).\(\bar{a}\)
Solution:
\(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are unit vectors

Adding (2), (3), (4) and using the fact that scalar product commutative, we get

Question 23.
If a parallelogram is constructed on the vectors \(\bar{a}=3 \bar{p}-\bar{q}\), \(\bar{b}=\bar{p}+3 \bar{q}\) and \(|\bar{p}|=|\bar{q}|=2\) and angle between \(\bar{p}\) and \(\bar{q}\) is\(\frac{\pi}{3}\) show that the ratio of the lengths of the sides is \(\sqrt {7}\) : \(\sqrt {13}\)
Solution:


Hence, the ratio of the lengths of the sides is \(\sqrt {7}\) : \(\sqrt {13}\).

Question 24.
Express the vector \(\bar{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}\) as a sum of two vectors such that one is parallel to the vector \(\bar{b}=3 \hat{i}+\hat{k}\) and other is perpendicular to \(\bar{b}\).
Solution:

By equality of vectors
3m + x = 5 … (1)
y = -2
and m – 3x = 5
From (1) and (2)
3m + x = m – 3x
∴ 2m = -4x m ∴ m = -2x
Substituting m = -2x in (1), we get
∴ -6x + x = 5 ∴ -5x = 5 ∴ x = -1
∴ m = -2x = 2

Question 25.
Find two unit vectors each of which makes equal angles with \(\bar{u}\), \(\bar{v}\) and \(\bar{w}\). \(\bar{u}=2 \hat{i}+\hat{j}-2 \hat{k}\), \(\bar{v}=\hat{i}+2 \hat{j}-2 \hat{k}\) and \(\bar{W}=2 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:

Question 26.
Find the acute angles between the curves at their points of intersection. y = x², y = x³
Solution:
The angle between the curves is same as the angle between their tangents at the points of intersection. We find the points of intersection of y = x² … (1)
and y = x³ … (2)
From (1) and (2)
x³ = x²
∴ x³ – x² = 0
∴ x²(x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0.
When x = 1, y = 1.

∴ equation of tangent to y = x³ at P is y = 0.
∴ the tangents to both curves at (0, 0) are y = 0
∴ angle between them is 0.
Angle at P = (1, 1)
Slope of tangent to y = x² at P

∴ equation of tangent to y = x³ at P is y – 1 = 3(x – 1) y = 3x – 2
We have to find angle between y = 2x – 1 and y = 3x – 2
Lines through origin parallel to these tagents are y = 2x and y = 3x
∴ \(\frac{x}{1}=\frac{y}{2}\) and \(\frac{x}{1}=\frac{y}{3}\)
These lines lie in XY-plane.
∴ the direction ratios of these lines are 1, 2, 0 and 1, 3, 0.
The angle θ between them is given by

Question 27.
Find the direction cosines and direction angles of the vector.
(i) \(2 \hat{i}+\hat{j}+2 \hat{k}\)
Solution:
Let \(\bar{a}\) = \(2 \hat{i}+\hat{j}+2 \hat{k}\)

(ii) \((1 / 2) \hat{i}+\hat{j}+\hat{k}\)
Solution:

Question 28.
Let \(\bar{b}\) = \(4 \hat{i}+3 \hat{j}\) and \(\bar{c}\) be two vectors perpendicular to each other in the XY-plane. Find vectors in the same plane having projection 1 and 2 along \(\bar{b}\) and \(\bar{c}\), respectively, are given y.
Solution:
\(\bar{b}\) = \(4 \hat{i}+3 \hat{j}\)

Question 29.
Show that no line in space can make angle \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X- axis and Y-axis.
Solution:
Let, if possible, a line in space make angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X-axis and Y-axis.

∴ cos²γ = 1 – \(\frac{3}{4}-\frac{1}{2}=-\frac{1}{4}\)
This is not possible, because cos γ is real
∴ cos²γ cannot be negative.
Hence, there is no line in space which makes angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X-axis and Y-axis.

Question 30.
Find the angle between the lines whose direction cosines are given by the equation 6mn – 2nl + 5lm = 0, 3l + m + 5n = 0
Solution:
Given 6mn – 2nl + 5lm = o
3l + m +5n = 0.
From (2), m = 3l – 5n
Putting the value of m in equation (1), we get,
⇒ 6n(-3l – 5n) – 2nl + 5l(-3l – 5n) = 0
⇒ -18nl- 30n – 2nl- 15l2 – 25nl = 0
⇒ – 30n² – 45nl – 15l² = 0
⇒ 2n² + 3nl + l² = 0
⇒ 2n² + 2nl + nl + l² = 0
⇒ (2n + l) (n + l) = 0
∴ 2n + l = 0 OR n + l = 0
∴ l = -2n OR l = -n
∴ l = -2n
From (2), 3l + m + 5n = 0
∴ -6n + m + 5n = 0
∴ m = n
i.e. (-2n, n, n) = (-2, 1, 1)
∴ l = -n
∴ -3n + m + 5n = 0
∴ m = -2n
i.e. (-n, -2n, n) = (1, 2, -1)
(a₁, b₁, c₁) = (-2, 1, 1) and (a₂, b₃, c₃) = (1, 2, -1)

Question 31.
If Q is the foot of the perpendicular from P(2, 4, 3) on the line joining the points A(1, 2, 4) and B(3, 4, 5), find coordinates of Q.
Solution:
Let PQ be the perpendicular drawn from point P(2, 4, 3) to the line joining the points A(1, 2, 4) and B (3, 4, 5).
Let Q divides AB internally in the ratio λ : 1

Now, direction ratios of AB are, 3 – 1, 4 – 2, 5 – 4 i.e., 2, 2, 1.

Coordinates of Q are,

Question 32.
Show that the area of a triangle ABC, the position vectors of whose vertices are a, b and c is \(\frac{1}{2}[\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}]\)
Question is modified.
Show that the area of a triangle ABC, the position vectors of whose vertices are \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) is \(\frac{1}{2}[\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}]\).
Solution:

Consider the triangle ABC.
Complete the parallelogram ABDC.
Vector area of ∆ABC

Question 33.
Find a unit vector perpendicular to the plane containing the point (a, 0, 0), (0, b, 0), and (0, 0, c). What is the area of the triangle with these vertices?
Solution:

Question 34.
State whether each expression is meaningful. If not, explain why ? If so, state whether it is a vector or a scalar.
(a) \(\bar{a} \cdot(\bar{b} \times \bar{c})\)
Solution:
This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(b) \(\bar{a} \times(\bar{b} \cdot \bar{c})\)
Solution:
This expression is meaningless because \(\bar{a}\) is a vector, \(\bar{b} \cdot \bar{c}\) is a scalar and vector product of vector and scalar is not defined.

(c) \(\bar{a} \times(\bar{b} \times \bar{c})\)
Solution:
This is vector product of two vectors. Therefore, this expression is meaningful and it is a vector.

(d) \(\bar{a} \cdot(\bar{b} \cdot \bar{c})\)
Solution:
This is meaningless because \(\bar{a}\) is a vector, \(\bar{b} \cdot \bar{c}\) is a scalar and scalar product of vector and scalar is not defined.

(e) \((\bar{a} \cdot \bar{b}) \times(\bar{c} \cdot \bar{d})\)
Solution:
This is meaningless because \(\bar{a} \cdot \bar{b}, \bar{c} \cdot \bar{d}\) are scalars and cross product of two scalars is not defined.

(f) \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})\)
Solution:
This is scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(g) \((\bar{a} \cdot \bar{b}) \cdot \bar{c}\)
Solution:
This is meaningless because \(\bar{c}\) is a vector, \(\bar{a} \cdot \bar{b}\) scalar and scalar product of vector and scalar is not defined.

(h) \((\bar{a} \cdot \bar{b}) \bar{c}\)
Solution:
This is a scalar multiplication of a vector. Therefore, this expression is meaningful and it is a vector.

(i) \((|\bar{a}|)(\bar{b} \cdot \bar{c})\)
Solution:
This is the product of two scalars. Therefore, this expression is meaningful and it is a scalar.

(j) \(\bar{a} \cdot(\bar{b}+\bar{c})\)
Solution:
This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(k) \(\bar{a} \cdot \bar{b}+\bar{c}\)
Solution:
This is the sum of scalar and vector which is not defined. Therefore, this expression is meaningless.

(l) \(|\bar{a}| \cdot(\bar{b}+\bar{c})\)
Solution:
This is meaningless because \(\bar{a}\) is a vector, \(\overline{\mathrm{b}}+\overline{\mathrm{c}}\) is a scalar and the scalar product of vector and scalar is not defined.

Question 35.
Show that, for any vectors \(\bar{a}, \bar{b}, \bar{c}\)
\((\bar{a}+\bar{b}+\bar{c}) \times \bar{c}+(\bar{a}+\bar{b}+\bar{c}) \times \bar{b}+(\bar{b}+\bar{c}) \times \bar{a}=2 \bar{a} \times \bar{c}\)
Question is modified.
For any vectors \(\bar{a}, \bar{b}, \bar{c}\) show that
\(\begin{aligned}
&(\bar{a}+\bar{b}+\bar{c}) \times \bar{c}+(\bar{a}+\bar{b}+\bar{c}) \times \bar{b}+(\bar{b}-\bar{c}) \times \bar{a} \\
&=2 \bar{a} \times \bar{c} .
\end{aligned}\)
Solution:

Question 36.
Suppose that \(\bar{a}\) = 0.
(a) If \(\bar{a} \cdot \bar{b}=\bar{a} \cdot \bar{c}\) then is \(\bar{b}=\bar{c}\)?
Solution:

(b) If \(\bar{a} \times \bar{b}=\bar{a} \times \bar{c}\) then is \(\bar{b}=\bar{c}\)?
Solution:

(c) If \(\bar{a} \cdot \bar{b}=\bar{a} \cdot \bar{c}\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{c}}\) then is \(\bar{b}=\bar{c}\)?
Solution:

Question 37.
If A(3, 2, -1), B(-2, 2, -3), C(3, 5, -2), D(-2, 5, -4) then
(i) verify that the points are the vertices of a parallelogram and
Solution:

∴ opposite sides AB and DC of ABCD are parallel and equal.
∴ ABCD is a parallelogram.

(ii) find its area.
Solution:

Question 38.
Let A, B, C, D be any four points in space. Prove that \(|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|\) = 4 (area of ∆ABC)
Solution:
Let A, B, C, D have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\) respectively.
Consider \(|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|\)

Question 39.
Let \(\hat{a}, \hat{b}, \hat{c}\) be unit vectors such that \(\hat{a} \cdot \hat{b}=\hat{a} \cdot \hat{c}=0\) and the angle between \(\hat{b}\) and \(\hat{c}\) be\(\frac{\pi}{6}\).
Prove that \(\hat{a}=\pm 2(\hat{b} \times \hat{c})\)
Solution:
\(\hat{a} \cdot \hat{b}=\hat{a} \cdot \hat{c}=0\)
∴ \(\hat{a}\) is perpendicular to \(\hat{b}\) and \(\hat{c}\) both
∴ \(\hat{a}\) is parallel to \(\hat{b}\) × \(\hat{c}\)
∴ \(\hat{a}\) = m(\(\hat{b}\) × \(\hat{c}\)), m is a scalar.

Question 40.
Find the value of ‘a’ so that the volume of parallelopiped a formed by \(\hat{i}+\hat{j}+\hat{k}+a \hat{k}\) aand \(a j+\hat{k}\) becomes minimum.
Question is modified.
Find the value of ‘a’ so that the volume of parallelopiped formed by \(\hat{i}+a \hat{j}+\hat{k}, \hat{j}+a \hat{k}\) and \(a \hat{i}+\hat{k}\) becomes minimum.
Solution:
Let \(\bar{p}\) = \(\hat{i}+a \hat{j}+\hat{k}\), \(\bar{q}\) = \(\hat{j}+a \hat{k}\), \(\bar{r}\) = \(a \hat{i}+\hat{k}\)
Let V be the volume of the parallelopiped formed by \(\bar{p}, \bar{q}, \bar{r}\).

Question 41.
Find the volume of the parallelepiped spanned by the diagonals of the three faces of a cube of side a that meet at one vertex of the cube.
Solution:

Take origin O as one vertex of the cube and OA, OB and OC as the positive directions of the X-axis, the Y-axis and the Z-axis respectively.
Here, the sides of the cube are
OA = OB = OC = a
∴ the coordinates of all the vertices of the cube will be
O = (0, 0, 0) A = (a, 0, 0)
B = (0, a, 0) C = (0, 0, a)
N = (a, a, 0) L = (0, a, a)
M = (a, 0, a) P = (a, a, a)
ON, OL, OM are the three diagonals which meet at the vertex O

Question 42.
If \(\bar{a}, \bar{b}, \bar{c}\) are three non-coplanar vectors, then show that \(\frac{\bar{a} \cdot(\bar{b} \times \bar{c})}{(\bar{c} \times \bar{a}) \cdot \bar{b}}+\frac{\bar{b} \cdot(\bar{a} \times \bar{c})}{(\bar{c} \times \bar{a}) \cdot \bar{b}}\) = 0
Solution:

Question 43.
Prove that \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})\left|\begin{array}{ll}
\bar{a} \cdot \bar{c} & \bar{b} \cdot \bar{c} \\
\bar{a} \cdot \bar{d} & \bar{b} \cdot \bar{d}
\end{array}\right|\)
Solution:

Question 44.
Find the volume of a parallelopiped whose coterminus edges are represented by the vector \(\hat{j}+\hat{k} \cdot \hat{i}+\hat{k}\) and \(\hat{i}+\hat{j}\). Also find volume of tetrahedron having these coterminous edges.
Solution:
Let \(\bar{a}\) = \(\hat{j}+\hat{k}\), \(\bar{b}\) = \(\hat{i}+\hat{k}\) and \(\bar{c}\) = \(\hat{i}+\hat{j}\) be the co-terminus edges of a parallelopiped.
Then volume of the parallelopiped = \([\bar{a} \bar{b} \bar{c}]\)
= \(\left|\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 0(0 – 1) – 1(0 – 1) + 1(1 – 0)
= 0 + 1 + 1 = 2cu units.
Also, volume of tetrahedron = \(\frac{1}{6}[\bar{a} \bar{b} \bar{c}]\)
= \(\frac{1}{6}(2)=\frac{1}{3}\) cubic units.

Question 45.
Using properties of scalar triple product, prove that \(\left[\begin{array}{llll}
\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}
\end{array}\right]=2\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\).
Solution:

Question 46.
If four points A(\(\bar{a}\)), B(\(\bar{b}\)), C(\(\bar{c}\)) and D(\(\bar{d}\)) are coplanar then show that \(\left[\begin{array}{lll}
\bar{a} \bar{b} \bar{d}]+\left[\begin{array}{lll}
\bar{b} & \bar{c} & \bar{d}
\end{array}\right]+\left[\begin{array}{lll}
\bar{c} & \bar{a} & \bar{d}
\end{array}\right]=[\overline{\bar{a}} \bar{b} & \bar{c}
\end{array}\right]\)
Solution:

Question 47.
If \(\bar{a}\) \(\bar{b}\) and \(\bar{c}\) are three non coplanar vectors, then \((\bar{a}+\bar{b}+\bar{c}) \cdot[(\bar{a}+\bar{b}) \times(\bar{a}+\bar{c})]=-\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\).
Solution:

Question 48.
If in a tetrahedron, edges in each of the two pairs of opposite edges are perpendicular, then show that the edges in the third pair are also perpendicular.
Solution:

Let O-ABC be a tetrahedron. Then o
(OA, BC), (OB, CA) and (OC, AB) are the pair of opposite edges.
Take O as the origin of reference and let \(\bar{a}\) \(\bar{b}\) and
∴ the third pair (OC, AB) is perpendicular.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.5

Question 1.
Find \(\bar{a}\)∙(\(\bar{b}\) × \(\bar{c}\)), if \(\bar{a}\) = \(3 \hat{i}-\hat{j}+4 \hat{k}\), \(\bar{b}\) = \(2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\bar{c}\) = \(-5 \hat{i}+2 \hat{j}+3 \hat{k}\)
Solution:
\(\bar{a}\)∙(\(\bar{b}\) × \(\bar{c}\)) = \(\left|\begin{array}{rrr}
3 & -1 & 4 \\
2 & 3 & -1 \\
-5 & 2 & 3
\end{array}\right|\)
= 3(9 + 2) + 1 (6 – 5) + 4(4 + 15)
= 33 + 1 + 76
= 110.

Question 2.
If the vectors \(3 \hat{i}+5 \hat{k}, 4 \hat{i}+2 \hat{j}-3 \hat{k}\) and \(3 \hat{i}+\hat{j}+4 \hat{k}\) are to co-terminus edges of the parallelo piped, then find the volume of the parallelopiped.
Solution:

= 3(8 + 3) – 0(16 + 9) + 5(4 – 6)
= 33 – 0 – 10 = 23
∴ volume of the parallelopiped = \([\bar{a} \bar{b} \bar{c}]\)
= 23 cubic units.

Question 3.
If the vectors \(-3 \hat{i}+4 \hat{j}-2 \hat{k}, \hat{i}+2 \hat{k}\) and\(\hat{i}-p \hat{j}\) are coplanar, then find the value of p.
Solution:

∴ -3(0 + 2p) – 4(0 – 2) – 2(-p – 0) = 0
∴ -6p + 8 + 2p = 0
∴ -4p = -8
P = 2.

Question 4.
Prove that :
(i) [latex]\bar{a} \bar{b}+\bar{c} \bar{a}+\bar{b}+\bar{c}[/latex] = 0
Solution:

(ii) (\(\bar{a}\) – \(2 \bar{b}\) – \(\bar{c}\))∙[(\(\bar{a}\) – \(\bar{b}\)) × \(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)] = 3[\(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)]
Question is modified.
(\(\bar{a}\) – \(2 \bar{b}\) – \(\bar{c}\)) [(\(\bar{a}\) – \(\bar{b}\)) × \(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)] = 3[\(\bar{a}\) \(\bar{b}\) \(\bar{c}\)]
Solution:

Question 5.
If \(\bar{c}\) =3\(\bar{a}\) – 2\(\bar{b}\) prove that [\(\bar{a}\) \(\bar{b}\) \(\bar{c}\)] = 0
Solution:
We use the results :\(\bar{b}\) × \(\bar{b}\) = 0 and if in a scalar triple product, two vectors are equal, then the scalar triple product is zero.

Question 6.
If u = \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{k}\), \(\bar{v}\) = \(3 \hat{\mathbf{i}}+\hat{k}\) and \(\bar{w}\) = \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\) are given vectors, then find
(i) [\(\bar{u}\) + \(\bar{w}\)]∙[(\(\bar{w}\) × \(\bar{r}\)) × (\(\bar{r}\) × \(\bar{w}\))]
Question is modified.
If \(\bar{u}\) = \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{k}\), \(\bar{r}\) = \(3 \hat{\mathbf{i}}+\hat{k}\) and \(\bar{w}\) = \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\) are given vectors, then find [\(\bar{u}\) + \(\bar{w}\)]∙[(\(\bar{u}\) × \(\bar{r}\)) × (\(\bar{r}\) × \(\bar{w}\))]
Solution:

= 1(6 – 18) + 1 (-6 + 6) + 0
= -12 + 0 + 0 = -12.

Question 7.
Find the volume of a tetrahedron whose vertices are A( -1, 2, 3) B(3, -2, 1), C (2, 1, 3) and D(-1, -2, 4).
Solution:
The position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d }\) of the points A, B, C and D w.r.t. the origin are \(\bar{a}\) = \(-\hat{i}+2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(3 \hat{i}-2 \hat{j}+\hat{k}\), \(\bar{c}\) = \(2 \hat{i}+\hat{j}+3 \hat{k}\) and

Question 8.
If \(\bar{a}\) = \(\hat{i}+2 \hat{j}+3\), \(\bar{b}\) = \(3 \hat{i}+2 \hat{j}\) and \(\bar{c}\) = ,\(2 \hat{i}+\hat{j}+3\) then verify that \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) = (\(\bar{a}\) ⋅ \(\bar{c}\))\(\bar{b}\) – (\(\bar{a}\) ⋅ \(\bar{b}\))\(\bar{c}\)
Solution:


From (1) and (2), we get
\(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) = (\(\bar{a}\) ⋅ \(\bar{c}\))\(\bar{b}\) – (\(\bar{a}\) ⋅ \(\bar{b}\))\(\bar{c}\)

Question 9.
If, \(\bar{a}\) = \(\hat{i}-2 \hat{j}\), \(\bar{b}\) = \(\hat{i}+2 \hat{j}\) and \(\bar{c}\) =\(2 \hat{i}+\hat{j}-2\) then find
(i) \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\))
Solution:

(ii) (\(\bar{a}\) × \(\bar{b}\)) × \(\bar{c}\) Are the results same? Justify.
Solution:

\(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) ≠ (\(\bar{a}\) × \(\bar{b}\)) × \(\bar{c}\)

Question 10.
Show that \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) + \(\bar{b}\) × (\(\bar{c}\) × \(\bar{a}\)) + \(\bar{c}\) × (\(\bar{a}\) × \(\bar{b}\)) = 0
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.4

Question 1.
If \(\bar{a}\) = \(2 \hat{i}+3 \hat{j}-\hat{k}\), \(\bar{b}\) = \(\hat{i}-4 \hat{j}+2 \hat{k}\) find (\(\bar{a}\) + \(\bar{b}\)) × (\(\bar{a}\) – \(\bar{b}\))
Solution:

Question 2.
Find a unit vector perpendicular to the vectors \(\hat{j}+2 \hat{k}\) and \(\hat{i}+\hat{j}\).
Solution:
Let \(\bar{a}\) = \(\hat{j}+2 \hat{k}\), \(\bar{b}\) = \(\hat{i}+\hat{j}\)

Question 3.
If \(\bar{a} \cdot \bar{b}\) = \(\sqrt {3}\) and \(\bar{a} \times \bar{b}\) = \(2 \hat{i}+\hat{j}+2 \hat{k}\), find the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:
Let θ be the angle between \(\bar{a}\) and \(\bar{b}\)

∴ θ = 60°.

Question 4.
If \(\bar{a}\) = \(2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\bar{b}\) = \(\hat{i}-2 \hat{j}+\hat{k}\), find a vector of magnitude 5 perpendicular to both \(\bar{a}\) and \(\bar{b}\).
Solution:
Given : \(\bar{a}\) = \(2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\bar{b}\) = \(\hat{i}-2 \hat{j}+\hat{k}\)

∴ unit vectors perpendicular to both the vectors \(\bar{a}\) and \(\bar{b}\).

∴ required vectors of magnitude 5 units
= ±\(\frac{5}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)

Question 5.
Find
(i) \(\bar{u}\)∙\(\bar{v}\) if \(|\bar{u}|\) = 2, \(|\vec{v}|\) = 5, \(|\bar{u} \times \bar{v}|\) = 8
Solution:
Let θ be the angle between \(\bar{u}\) and \(\bar{v}\).
Then \(|\bar{u} \times \bar{v}|\) = 8 gives
\(|\bar{u}||\bar{v}|\) sin θ = 8
∴ 2 × 5 × sin θ = 8

(ii) \(|\bar{u} \times \bar{v}|\) if \(|\bar{u}|\) = 10, \(|\vec{v}|\) = 2, \(\bar{u} \cdot \bar{v}\) = 12
Solution:
Let θ be the angle between \(\bar{u}\) and \(\bar{v}\).
Then \(\bar{u} \cdot \bar{v}\) = 12 gives
\(|\bar{u} \| \bar{v}|\)cos θ = 12
∴ 10 × 2 × cos θ = 12

Question 6.
Prove that 2(\(\bar{a}\) – \(\bar{b}\)) × 2(\(\bar{a}\) + \(\bar{b}\)) = 8(\(\bar{a}\) × \(\bar{b}\))
Solution:

Question 7.
If \(\bar{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(4 \hat{i}-3 \hat{j}+\hat{k}\), and \(\bar{c}\) = \(\hat{i}-\hat{j}+2 \hat{k}\), verify that \(\bar{a}\) × (\(\bar{b}\) + \(\bar{c}\)) = \(\bar{a}\) × \(\bar{b}\) + \(\bar{a}\) × \(\bar{c}\)
Solution:
Given : \(\bar{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(4 \hat{i}-3 \hat{j}+\hat{k}\), \(\bar{c}\) = \(\hat{i}-\hat{j}+2 \hat{k}\)

Question 8.
Find the area of the parallelogram whose adjacent sides are the vectors \(\bar{a}\) = \(2 \hat{i}-2 \hat{j}+\hat{k}\) and \(\bar{b}\) = \(\hat{i}-3 \hat{j}-3 \hat{k}\).
Solution:
Given : \(\bar{a}\) = \(2 \hat{i}-2 \hat{j}+\hat{k}\), \(\bar{b}\) = \(\hat{i}-3 \hat{j}-3 \hat{k}\)

Area of the parallelogram whose adjacent sides are \(\bar{a}\) and \(\bar{b}\) is \(|\bar{a} \times \bar{b}|\) =\(\sqrt {146}\) sq units.

Question 9.
Show that vector area of a quadrilateral ABCD is \(\frac{1}{2}\) (\(\overline{A C}\) × \(\overline{B D}\)), where AC and BD are its diagonals.
Solution:
Let ABCD be a parallelogram.
Then \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BC}}\)

Question 10.
Find the area of parallelogram whose diagonals are determined by the vectors \(\bar{a}\) = \(3 i-\hat{j}-2 \hat{k}\), and \(\bar{b}\) = \(-\hat{i}+3 \hat{j}-3 \hat{k}\)
Solution:
Given: \(\bar{a}\) = \(3 i-\hat{j}-2 \hat{k}\), \(\bar{b}\) = \(-\hat{i}+3 \hat{j}-3 \hat{k}\)

Question 11.
If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) are four distinct vectors such that \(\bar{a} \times \bar{b}=\bar{c} \times \bar{d}\) and \(\bar{a} \times \bar{c}=\bar{b} \times \bar{d}\), prove that \(\bar{a}\) – \(\bar{d}\) is parallel to \(\bar{b}\) – \(\bar{c}\).
Solution:
\(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) are four distinct vectors

Question 12.
If \(\bar{a}\) = \(\hat{i}+\hat{j}+\hat{k}\) and, \(\bar{c}\) = \(\hat{j}-\hat{k}\), find a vector \(\bar{b}\) satisfying \(\bar{a}\) × \(\bar{b}\) = \(\bar{c}\) and \(\bar{a} \cdot \bar{b}\) = 3
Solution:
Given \(\bar{a}\) = \(\hat{i}+\hat{j}+\hat{k}\), \(\bar{c}\) = \(\hat{j}-\hat{k}\)

By equality of vectors,
z – y = 0 ….(2)
x – z = 1 ……(3)
y – x = -1 ……(4)
From (2), y = z.
From (3), x = 1 + z
Substituting these values of x and y in (1), we get
1 + z + z + z = 3 ∴ z = \(\frac{2}{3}\)
∴ y = z = \(\frac{2}{3}\)
∴ x = 1 + z =1 + \(\frac{2}{3}=\frac{5}{3}\)

Question 13.
Find \(\bar{a}\), if \(\bar{a} \times \hat{i}+2 \bar{a}-5 \hat{j}=\overline{0}\).
Solution:

By equality of vectors
2x = 0 i.e. x = 0
2y + z – 5 = 0 … (1)
2z – y = 0 … (2)
From (2), y = 2z
Substituting y = 2z in (1), we get
4z + z = 5 ∴ z = 1
∴ y = 2z = 2(1) = 2
∴ x = 0, y = 2, z = 1
∴ \(\bar{a}=2 \hat{j}+\hat{k}\)

Question 14.
If \(|\bar{a} \cdot \bar{b}|\) = \(|\bar{a} \times \bar{b}|\) and \(\bar{a} \cdot \bar{b}\) < 0, then find the angle between \(\bar{a}\) and \(\bar{b}\)
Solution:
Let θ be the angle between \(\bar{a}\) and \(\bar{b}\).
Then \(|\bar{a} \cdot \bar{b}|\) = \(|\bar{a} \times \bar{b}|\) gives

Hence, the angle between \(\bar{a}\) and \(\bar{b}\) is \(\frac{3 \pi}{4}\).

Question 15.
Prove by vector method that sin (α + β) = sinα∙cosβ+cosα∙sinβ.
Solution:

Let ∠XOP and ∠XOQ be in standard position and m∠XOP = -α, m∠XOQ = β.
Take a point A on ray OP and a point B on ray OQ such that
OA = OB = 1.
Since cos (-α) = cos α
and sin (-α) = -sin α,
A is (cos (-α), sin (-α)),
i.e. (cos α, – sin α)
B is (cos β, sin β)

The angle between \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is α + β.
Also \(\overline{\mathrm{OA}}\), \(\overline{\mathrm{OB}}\) lie in the XY-plane.
∴ the unit vector perpendicular to \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is \(\bar{k}\).
∴ \(\overline{\mathrm{OA}}\) × \(\overline{\mathrm{OB}}\) = [OA∙OB sin (α + β)]\(\bar{k}\)
= sin(α + β)∙\(\bar{k}\) …(2)
∴ from (1) and (2),
sin (α + β) = sin α cos β + cos α sin β.

Question 16.
Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are
(i) -2, 1, -1 and -3, -4, 1
Solution:
Let a, b, c be the direction ratios of the vector which is perpendicular to the two lines whose direction ratios are -2, 1, -1 and -3, -4, 1
∴ -2a + b – c = 0 and -3a – 4b + c = 0

∴ the required direction ratios are -3, 5, 11
Alternative Method:
Let \(\bar{a}\) and \(\bar{b}\) be the vectors along the lines whose direction ratios are -2, 1, -1 and -3, -4, 1 respectively.
Then \(\bar{a}\) = \(-2 \hat{i}+\hat{j}-\hat{k}\) and \(\bar{b}\) = \(-3 \hat{i}-4 \hat{j}+\hat{k}\)
The vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is given by

Hence, the required direction ratios are -3, 5, 11.

(ii) 1, 3, 2 and -1, 1, 2
Solution:

Question 17.
Prove that two vectors whose direction cosines are given by relations al + bm + cn = 0 and fmn + gnl + hlm = 0 are perpendicular if \(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\) = 0
Solution:
Given, al + bm + cn = 0 …(1)
and fmn + gnl + hlm = 0 …..(2)
From (1), n = \(-\left(\frac{a l+b m}{c}\right)\) …..(3)
Substituting this value of n in equation (2), we get
(fm + gl)∙[latex]-\left(\frac{a l+b m}{c}\right)[/latex] + hlm = 0
∴ -(aflm + bfm² + agl² + bglm) + chlm = 0
∴ agl² + (af + bg – ch)lm + bfm² = 0 … (4)
Note that both l and m cannot be zero, because if l = m = 0, then from (3), we get
n = 0, which is not possible as l² + m² + n² = 1.
Let us take m # 0.
Dividing equation (4) by m², we get
ag\(\left(\frac{l}{m}\right)^{2}\) + (af + bg – ch)\(\left(\frac{l}{m}\right)\) + bf = 0 … (5)
This is quadratic equation in \(\left(\frac{l}{m}\right)\).
If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of the two lines given by the equation (1) and (2), then \(\frac{l_{1}}{m_{1}}\) and \(\frac{l_{2}}{m_{2}}\) are the roots of the equation (5).
From the quadratic equation (5), we get

i.e. if \(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\) = 0.

Question 18.
If A(1, 2, 3) and B(4, 5, 6) are two points, then find the foot of the perpendicular from the point B to the line joining the origin and point A.
Solution:

Let M be the foot of the perpendicular drawn from B to the line joining O and A.
Let M = (x, y, z)
OM has direction ratios x – 0, y – 0, z – 0 = x, y, z
OA has direction ratios 1 – 0, 2 – 0, 3 – 0 = 1, 2, 3
But O, M, A are collinear.
∴ \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) = k …(Let)
∴ x = k, y = 2k, z = 3k
∴ M = (k, 2k, 3k)
∵ BM has direction ratios
k – 4, 2k – 5, 3k – 6
BM is perpendicular to OA
∴ (l)(k – 4) + 2(2k – 5) + 3(3k – 6)
∴ = k – 4 + 4k – 10 + 9k – 18 = 0
∴ 14k = 32
∴ k = \(\frac{16}{7}\)
∴ M = (k, 2k, 3k) = (\(\frac{16}{7}, \frac{32}{7}, \frac{48}{7}\))

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.3

Question 1.
Find two unit vectors each of which is perpendicular to both
\(\bar{u}\) and \(\bar{v}\), where \(\bar{u}\) = \(2 \hat{i}+\hat{j}-2 \hat{k}\), \(\bar{v}\) = \(\hat{i}+2 \hat{j}-2 \hat{k}\)
Solution:
Let \(\bar{u}\) = \(2 \hat{i}+\hat{j}-2 \hat{k}\)
\(\bar{v}\) = \(\hat{i}+2 \hat{j}-2 \hat{k}\)

Question 2.
If \(\bar{a}\) and \(\bar{b}\) are two vectors perpendicular to each other, prove that (\(\bar{a}\) + \(\bar{b}\))² = (\(\bar{a}\) – \(\bar{b}\))²
Solution:
\(\bar{a}\) and \(\bar{b}\) are perpendicular to each other.

∴ LHS = RHS
Hence, (\(\bar{a}\) + \(\bar{b}\))² = (\(\bar{a}\) – \(\bar{b}\))².

Question 3.
Find the values of c so that for all real x the vectors \(x c \hat{i}-6 \hat{j}+3 \hat{k}\) and \(x \hat{i}+2 \hat{j}+2 c x \hat{k}\) make an obtuse angle.
Solution:
Let \(\bar{a}\) = \(x c \hat{i}-6 \hat{j}+3 \hat{k}\) and \(\bar{b}\) = \(x \hat{i}+2 \hat{j}+2 c x \hat{k}\)
Consider \(\bar{a}\)∙\(\bar{b}\) = \((x c \hat{i}-6 \hat{j}+3 \hat{k}) \cdot(x \hat{i}+2 \hat{j}+2 c x \hat{k})\)
= (xc)(x) + (-6)(2) + (3)(2cx)
= cx² – 12 + 6cx
= cx² + 6cx – 12
If the angle between \(\bar{a}\) and \(\bar{b}\) is obtuse, \(\bar{a}\)∙\(\bar{b}\) < 0.
∴ cx² + 6cx – 12 < 0
∴ cx² + 6cx < 12

∴ c < 0.
Hence, the angle between a and b is obtuse if c < 0.

Question 4.
Show that the sum of the length of projections of \(\hat{p} \hat{i}+q \hat{j}+r \hat{k}\) on the coordinate axes, where p = 2, q = 3 and r = 4, is 9.
Solution:
Let \(\bar{a}\) = \(\hat{p} \hat{i}+q \hat{j}+r \hat{k}\)
Projection of \(\bar{a}\) on X-axis

Similarly, projections of \(\bar{a}\) on Y- and Z-axes are 3 and 4 respectively.
∴ sum of these projections = 2 + 3 + 4 = 9.

Question 5.
Suppose that all sides of a quadrilateral are equal in length and opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular.
Solution:


∵ \(\overline{\mathrm{AC}}\), \(\overline{\mathrm{BD}}\) are non-zero vectors
∴ \(\overline{\mathrm{AC}}\) is perpendicular to \(\overline{\mathrm{BD}}\)
Hence, the diagonals are perpendicular.

Question 6.
Determine whether \(\bar{a}\) and \(\bar{b}\) are orthogonal, parallel or neither.
(i) \(\bar{a}\) = \(-9 \hat{i}+6 \hat{j}+15 \hat{k}\), \(\bar{b}\) = \(6 \hat{i}-4 \hat{j}-10 \hat{k}\)
Solution:
\(\bar{a}\) = \(-9 \hat{i}+6 \hat{j}+15 \hat{k}\) = -3\((3 \hat{i}-2 \hat{j}-5 \hat{k})\)
= \(-\frac{3}{2}(6 \hat{i}-4 \widehat{j}-19 \hat{k})\)
∴ \(\bar{a}\) = \(-\frac{3}{2} \bar{b}\)
i.e. \(\bar{a}\) is a non-zero scalar multiple of \(\bar{b}\)
Hence, \(\bar{a}\) is parallel to \(\bar{b}\).

(ii) \(\bar{a}\) = \(2 \hat{i}+3 \hat{j}-\hat{k}\), \(\bar{b}\) = \(5 \hat{i}-2 \hat{j}+4 \hat{k}\)
Solution:
\(\bar{a} \cdot \bar{b}\) = \((2 \hat{i}+3 \hat{j}-\hat{k}) \cdot(5 \hat{i}-2 \hat{j}+4 \hat{k})\)
= (2)(5) + (3)(-2) + (-1)(4)
= 10 – 6 – 4 = 0
Since, \(\bar{a}\), \(\bar{b}\) are non-zero vectors and \(\bar{a} \cdot \bar{b}\) = 0, \(\bar{a}\) is orthogonal to \(\bar{b}\).

(iii) \(\bar{a}\) = \(-\frac{3}{5} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{3} \hat{k}\), \(\bar{b}\) = \(5 \hat{i}+4 \hat{j}+3 \hat{k}\).
Solution:

= -3 + 2 + 1
= 0
Since, \(\bar{a}\), \(\bar{b}\) are non-zero vectors and \(\bar{a} \cdot \bar{b}\) = 0
\(\bar{a}\) is orthogonal to \(\bar{b}\).

(iv) \(\bar{a}\) = \(4 \hat{i}-\hat{j}+6 \hat{k}\), \(\bar{a}\) = \(5 \hat{i}-2 \hat{j}+4 \hat{k}\)
Solution:
\(\bar{a} \cdot \bar{b}\) = \((4 \hat{i}-\hat{j}+6 \hat{k}) \cdot(5 \hat{i}-2 \hat{j}+4 \hat{k})\)
= (4)(5) + (-1)(-2) + (6)(4)
= 20 + 2 + 24
= 46 ≠ 0
∴ \(\bar{a}\) is not orthogonal to \(\bar{b}\).
It is clear that \(\bar{a}\) is not a scalar multiple of \(\bar{b}\).
∴ \(\bar{a}\) is not parallel to \(\bar{b}\).
Hence, \(\bar{a}\) is neither parallel nor orthogonal to \(\bar{b}\).

Question 7.
Find the angle P of the triangle whose vertices are P(0, -1, -2), Q(3, 1, 4) and R(5, 7, 1).
Solution:
The position vectors \(\bar{p}\), \(\bar{q}\) and \(\bar{r}\) of the points P(0, -1, -2), Q(3, 1, 4) and R(5, 7, 1) are


∴ P = 45°

Question 8.
If\(\hat{p}\), \(\hat{q}\), and \(\hat{r}\) are unit vectors, find

(i) \(\hat{p} \cdot \hat{q}\)
Solution:

Let the triangle be denoted by ABC,
where \(\overline{\mathrm{AB}}\) = \(\bar{p}\), \(\overline{\mathrm{AC}}\) = \(\bar{q}\) and \(\overline{\mathrm{BC}}\) = \(\bar{r}\)
∵ \(\bar{p}\), \(\bar{r}\), \(\bar{r}\) are unit vectors.
∴ l(AB) = l(BC) = l(CA) = 1
∴ the triangle is equilateral
∴ ∠A = ∠B = ∠C = 60°
(i) Using the formula for angle between two vectors,

(ii) \(\hat{p} \cdot \hat{r}\)
Solution:

Question 9.
Prove by vector method that the angle subtended on semicircle is a right angle.
Solution:
Let seg AB be a diameter of a circle with centre C and P be any point on the circle other than A and B. Then ∠APB is an angle subtended on a semicircle.
Let \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{CB}}\) = \(\bar{a}\) and \(\overline{\mathrm{CP}}\) = \(\bar{r}\).
Then|\(\bar{a}\)| = |\(\bar{r}\)| …(1)

Hence, the angle subtended on a semicircle is the right angle.

Question 10.
If a vector has direction angles 45ºand 60º find the third direction angle.
Solution:
Let α = 45°, β = 60°
We have to find γ.
∴ cos²α + cos²β + cos²γ = 1
∴ cos²45° + cos²60° + cos²γ = 1

Hence, the third direction angle is \(\frac{\pi}{3}\) or \(\frac{2 \pi}{3}\).

Question 11.
If a line makes angles 90º, 135º, 45º with the X, Y and Z axes respectively, then find its direction cosines.
Solution:
Let l, m, n be the direction cosines of the line.
Then l = cos α, m = cos β, n = cos γ
Here, α = 90°, β = 135° and γ = 45°
∴ l = cos 90° = 0
m = cos 135° = cos (180° – 45°) = -cos 45° = \(-\frac{1}{\sqrt{2}}\) and n = cos 45° = \(\frac{1}{\sqrt{2}}\)
∴ the direction cosines of the line are 0, \(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\).

Question 12.
If a line has the direction ratios, 4, -12, 18 then find its direction cosines.
Solution:
The direction ratios of the line are a = 4, b = -12, c = 18.
Let l, m, n be the direction cosines of the line.

Question 13.
The direction ratios of \(\overline{A B}\) are -2, 2, 1. If A = (4, 1, 5) and l(AB) = 6 units, find B.
Solution:
The direction ratio of \(\overline{A B}\) are -2, 2, 1.
∴ the direction cosines of \(\overline{A B}\) are

The coordinates of the points which are at a distance of d units from the point (x₁, y₁, z₁) are given by (x₁ ± ld, y₁ ± md, z₁ ± nd)
Here x₁ = 4, y₁ = 1, z₁ = 5, d = 6, l = \(-\frac{2}{3}\), m = \(\frac{2}{3}\), n = \(\frac{1}{3}\)
∴ the coordinates of the requited points are
(4 ± \(\left(-\frac{2}{3}\right)\)6, 1 ± \(\frac{2}{3}\)(6), 5 ± \(\frac{1}{3}\)(6))
i.e. (4 – 4, 1 + 4, 5 + 2) and (4 + 4, 1 – 4, 5 – 2)
i.e. (0, 5, 7) and (8, -3, 3).

Question 14.
Find the angle between the lines whose direction cosines l, m, n satisfy the equations 5l + m + 3n = 0 and 5mn – 2nl + 6lm = 0.
Solution:
Given, 5l + m + 3n = 0 …(1)
and 5mn – 2nl + 6lm = 0 …(2)
From (1), m = -(51 + 3n)
Putting the value of m in equation (2), we get,
-5(5l + 3n)n – 2nl – 6l(5l + 3n) = 0
∴ -25ln – 15n² – 2nl – 30l² – 18ln = 0
∴ – 30l² – 45ln – 15n² = 0
∴ 2l² + 3ln + n² = 0
∴ 2l² + 2ln + ln + n² = 0
∴ 2l(l + n) + n(l + n) = 0
∴ (l + n)(2l + n) = 0
∴ l + n = 0 or 2l + n = 0
l = -n or n = -2l
Now, m = -(5l + 3n), therefore, if l = -n,
m = -(-5n + 3n) = 2n
∴ -l = \(\frac{m}{2}\) = n
∴ \(\frac{l}{-1}=\frac{m}{2}=\frac{n}{1}\)
∴ the direction ratios of the first line are
a₁ = -1, b₁ = 2, c₁ = 1
If n = -2l, m = -(5l – 6l) – l
∴ l = m = \(\frac{n}{-2}\)
∴ \(\frac{l}{1}=\frac{m}{1}=\frac{n}{-2}\)
∴ the direction ratios of the second line are
a₂ = -1, b₂ = 1, c₂ = -2
Let θ be the angle between the lines.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.2

Question 1.
Find the position vector of point R which divides the line joining the points P and Q whose position vectors are \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(-5 \hat{i}+2 \hat{j}-5 \hat{k}\) in the ratio 3 : 2
(i) internally
Solution:
It is given that the points P and Q have position vectors \(\bar{p}\) = \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\bar{p}\) = \(-5 \hat{i}+2 \hat{j}-5 \hat{k}\) respectively.
(i) If R(\(\bar{r}\)) divides the line segment PQ internally in the ratio 3 : 2, by section formula for internal division,

(ii) externally.
Solution:
If R(\(\bar{r}\)) divides the line segment joining P and Q externally in the ratio 3 : 2, by section formula for external division,

∴ coordinates of R = (-19, 8, -21).
Hence, the position vector of R is \(-19 \hat{i}+8 \hat{j}-21 \hat{k}\) and coordinates of R are (-19, 8, -21).

Question 2.
Find the position vector of midpoint M joining the points L (7, -6, 12) and N (5, 4, -2).
Solution:
The position vectors \(\bar{l}\) and \(\bar{n}\) of the points L(7, -6, 12) and N (5, 4, -2) are given by

∴ coordinates of M = (6, -1, 5).
Hence, position vector of M is \(6 \hat{i}-\hat{j}+5 \hat{k}\) and the coordinates of M are (6, -1, 5).

Question 3.
If the points A(3, 0, p), B (-1, q, 3) and C(-3, 3, 0) are collinear, then find
(i) The ratio in which the point C divides the line segment AB.
Solution:
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be the position vectors of A, B and C respectively.
Then \(\bar{a}\) = \(3 \hat{i}+0 \cdot \hat{j}+p \hat{k}\), \(\bar{b}\) = \(-\hat{i}+q \hat{j}+3 \hat{k}\) and \(\bar{c}\) = \(-3 \hat{i}+3 \hat{j}+0 \hat{k}\).
As the points A, B, C are collinear, suppose the point C divides line segment AB in the ratio λ : 1.
∴ by the section formula,

By equality of vectors, we have,
-3(λ + 1) = -λ + 3 … (1)
3(λ + 1 ) = λ q … (2)
0 = 3λ + p … (3)
From equation (1), -3λ – 3 = -λ + 3
∴ -2λ = 6 ∴ λ = -3
∴ C divides segment AB externally in the ratio 3 : 1.

(ii) The values of p and q.
Solution:
Putting λ = -3 in equation (2), we get
3(-3 + 1) = -3q
∴ -6 = -3q ∴ q = 2
Also, putting λ = -3 in equation (3), we get
0 = -9 + p ∴ p = 9
Hence p = 9 and q = 2.

Question 4.
The position vector of points A and B are 6\(\bar{a}\) + 2\(\bar{b}\) and \(\bar{a}\) – 3\(\bar{b}\). If the point C divides AB in the ratio 3 : 2 then show that the position vector of C is 3\(\bar{a}\) – \(\bar{b}\).
Solution:
Let \(\bar{c}\) be the position vector of C.
Since C divides AB in the ratio 3 : 2,

Hence, the position vector of C is 3\(\bar{a}\) – \(\bar{b}\).

Question 5.
Prove that the line segments joining mid-point of adjacent sides of a quadrilateral form a parallelogram.
Solution:
Let ABCD be a quadrilateral and P, Q, R, S be the midpoints of the sides AB, BC, CD and DA respectively.
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\), \(\bar{p}\), \(\bar{q}\), \(\bar{r}\) and s be the position vectors of the points A, B, C, D, P, Q, R and S respectively.
Since P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively,


∴ □PQRS is a parallelogram.

Question 6.
D and E divide sides BC and CA of a triangle ABC in the ratio 2 : 3 respectively. Find the position vector of the point of intersection of AD and BE and the ratio in which this point divides AD and BE.
Solution:

Let AD and BE intersect at P.
Let A, B, C, D, E, P have position vectos \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\), \(\bar{e}\), \(\bar{p}\) respectively.
D and E divide segments BC and CA internally in the ratio 2 : 3.
By the section formula for internal division,

LHS is the position vector of the point which divides segment AD internally in the ratio 15 : 4.
RHS is the position vector of the point which divides segment BE internally in the ratio 10 : 9.
But P is the point of intersection of AD and BE.
∴ P divides AD internally in the ratio 15 : 4 and P divides BE internally in the ratio 10 : 9.
Hence, the position vector of the point of interaction of
AD and BE is \(\bar{p}\) = \(\frac{15 \bar{d}+4 \bar{a}}{19}=\frac{10 \bar{e}+9 \bar{b}}{19}\) and it divides AD internally in the ratio 15 : 4 and BE internally in the ratio 10 : 9.

Question 7.
Prove that a quadrilateral is a parallelogram if and only if its diagonals bisect each other.
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) be respectively the position vectors of the vertices A, B, C and D of the parallelogram ABCD. Then AB = DC and side AB || side DC.

The position vectors of the midpoints of the diagonals AC and BD are (\(\bar{a}\) + \(\bar{c}\))/2 and (\(\bar{b}\) + \(\bar{d}\))/2. By (1), they are equal.
∴ the midpoints of the diagonals AC and BD are the same.
This shows that the diagonals AC and BD bisect each other.

(ii) Conversely, suppose that the diagonals AC and BD
of □ ABCD bisect each other,
i. e. they have the same midpoint.
∴ the position vectors of these midpoints are equal.
∴ \(\frac{\bar{a}+\bar{c}}{2}=\frac{\bar{b}+\bar{d}}{2}\) ∴ \(\bar{a}+\bar{c}=\bar{b}+\bar{d}\)
∴ \(\bar{b}\) – \(\bar{a}\) = \(\bar{c}\) – \(\bar{d}\) ∴ \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{DC}}\)
∴ \(\overline{\mathrm{AB}}\) ||\(\overline{\mathrm{DC}}\) and \(|\overline{\mathrm{AB}}|\) = \(|\overline{\mathrm{DC}}|\)
∴ side AB || side DC and AB = DC.
∴ □ ABCD is a parallelogram.

Question 8.
Prove that the median of a trapezium is parallel to the parallel sides of the trapezium and its length is half the sum of parallel sides.
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) be respectively the position vectors of the vertices A, B, C and D of the trapezium ABCD, with side AD || side BC.
Then the vectors \(\overline{\mathrm{AD}}\) and \(\overline{\mathrm{BC}}\) are parallel.
∴ there exists a scalar k,
such that \(\overline{\mathrm{AD}}\) = k∙\(\overline{\mathrm{BC}}\)
∴ \(\overline{\mathrm{AD}}\) + \(\overline{\mathrm{BC}}\) = k∙\(\overline{\mathrm{BC}}\) + \(\overline{\mathrm{BC}}\)
= (k + 1)BC …(1)

Let \(\bar{m}\) and \(\bar{n}\) be the position vectors of the midpoints M and N of the non-parallel sides AB and DC respectively.
Then seg MN is the median of the trapezium.
By the midpoint formula,

Thus \(\overline{\mathrm{MN}}\) is a scalar multiple of \(\overline{\mathrm{BC}}\)
∴ \(\overline{\mathrm{MN}}\) and \(\overline{\mathrm{BC}}\) are parallel vectors
∴ \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{BC}}\) where \(\overline{\mathrm{BC}}\) || \(\overline{\mathrm{AD}}\)
∴ the median MN is parallel to the parallel sides AD and BC of the trapezium.
Now \(\overline{\mathrm{AD}}\) and \(\overline{\mathrm{BC}}\) are collinear

Question 9.
If two of the vertices of the triangle are A(3, 1, 4) and B(-4, 5, -3) and the centroid of a triangle is G(-1, 2, 1), then find the coordinates of the third vertex C of the triangle.
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{g}\) be the position vectors of A, B, C and G respectively.
Then \(\bar{a}\) = \(3 \hat{i}+\hat{j}+4 \hat{k}\), \(\bar{b}\) = \(-4 \hat{i}+5 \hat{j}-3 \hat{k}\) and \(\bar{g}\) = \(-\hat{i}+2 \hat{j}+\hat{k}\).
Since G is the centroid of the ∆ABC, by the centroid formula,

∴ the coordinates of third vertex C are (-2, 0, 2).

Question 10.
In∆OAB, E is the mid-point of OB and D is the point on AB such that AD : DB = 2 : 1.
If OD and AE intersect at P, then determine the ratio OP : PD using vector methods.
Solution:

Let A, B, D, E, P have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{d}\), \(\bar{e}\), \(\bar{p}\) respectively w.r.t. O.
∵ AD : DB = 2 : 1.
∴ D divides AB internally in the ratio 2 : 1.
Using section formula for internal division, we get

LHS is the position vector of the point which divides OD internally in the ratio 3 : 2.
RHS is the position vector of the point which divides AE internally in the ratio 4 : 1.
But OD and AE intersect at P
∴ P divides OD internally in the ratio 3 : 2.
Hence, OP : PD = 3 : 2.

Question 11.
If the centroid of a tetrahedron OABC is (1, 2, -1) where A = (a, 2, 3), B = (1, b, 2), C = (2, 1, c) respectively, find the distance of P (a, b, c) from the origin.
Solution:
Let G = (1, 2, -1) be the centroid of the tetrahedron OABC.
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{g}\) be the position vectors of the points A, B, C, G respectively w.r.t. O.

By equality of vectors
a + 3 = 4, b + 3 = 8, c + 5= -4
∴ a = 1, b = 5, c = -9
∴ P = (a, b, c) = (1, 5, -9)
Distance of P from origion = \(\sqrt{1^{2}+5^{2}+(-9)^{2}}\)
= \(\sqrt{1+25+81}\)
= \(\sqrt{107}\)

Question 12.
Find the centroid of tetrahedron with vertices K(5, -7, 0), L(1, 5, 3), M(4, -6, 3), N(6, -4, 2) ?
Solution:
Let \(\bar{p}\), \(\bar{l}\), \(\bar{m}\), \(\bar{n}\) be the position vectors of the points K, L, M, N respectively w.r.t. the origin O.

Hence, the centroid of the tetrahedron is G = (4, -3, 2).