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

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

Question 1.

If the line \(\frac{x}{3}=\frac{y}{4}\) = z is perpendicular to the line \(\frac{x-1}{k}=\frac{y+2}{3}=\frac{z-3}{k-1}\) then the value of k is:

Solution:

(b) \(-\frac{11}{4}\)

Question 2.

The vector equation of line 2x – 1 = 3y + 2 = z – 2 is

Solution:

(a) \(\bar{r}=\left(\frac{1}{2} \hat{i}-\frac{2}{3} \hat{j}+2 \hat{k}\right)+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})\)

Question 3.

The direction ratios of the line which is perpendicular to the two lines \(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\) and \(\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}\) are

(A) 4, 5, 7

(B) 4, -5, 7

(C) 4, -5, -7

(D) -4, 5, 8

Solution:

(A) 4, 5, 7

Question 4.

The length of the perpendicular from (1, 6, 3) to the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\)

(A) 3

(B) \(\sqrt {11}\)

(C) \(\sqrt {13}\)

(D) 5

Solution:

(C ) \(\sqrt {13}\)

Question 5.

The shortest distance between the lines \(\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) is

Question is modified.

The shortest distance between the lines \(\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) is

Solution:

(c) \(\frac{3}{\sqrt{2}}\)

Question 6.

The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\). and coplanar if

(A) k = 1 or -1

(B) k = 0 or -3

(C) k = + 3

(D) k = 0 or -1

Solution:

(B ) k = 0 or -3

Question 7.

The lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{6}\) and are

(A) perpendicular

(B) inrersecting

(C) skew

(D) coincident

Solution:

(B) inrersecting

Question 8.

Equation of X-axis is

(A) x = y = z

(B) y = z

(C) y = 0, z = 0

(D) x = 0, y = 0

Solution:

(C) y = 0, z = 0

Question 9.

The angle between the lines 2x = 3y = -z and 6x = -y = -4z is

(A ) 45º

(B ) 30º

(C ) 0º

(D ) 90º

Solution:

(D ) 90º

Question 10.

The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are

(A ) 2, 1, 6

(B ) 2, 1, -6

(C ) 2, -1, 6

(D ) -2, 1, 6

Solution:

(B ) 2, 1, -6

Question 11.

The perpendicular distance of the plane 2x + 3y – z = k from the origin is \(\sqrt {14}\) units, the value

of k is

(A ) 14

(B ) 196

(C ) \(2\sqrt {14}\)

(D ) \(\frac{\sqrt{14}}{2}\)

Solution:

(A ) 14

Question 12.

The angle between the planes and \(\bar{r} \cdot(\bar{i}-2 \bar{j}+3 \bar{k})+4=0\) and \(\bar{r} \cdot(2 \bar{i}+\bar{j}-3 \bar{k})+7=0\) is

Solution:

(d) cos^{-1}\(\left(\frac{9}{14}\right)\)

Question 13.

If the planes \(\bar{r} \cdot(2 \bar{i}-\lambda \bar{j}+\bar{k})=3\) and \(\bar{r} \cdot(4 \bar{i}-\bar{j}+\mu \bar{k})=5\) are parallel, then the values of λ and μ are respectively.

Solution:

(d) \(\frac{1}{2}\), 2

Question 14.

The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is

(A ) x + y + z =1

(B ) x + y + z = 2

(C ) x + y + z = 3

(D ) x + y + z = 4

Solution:

(D ) x + y + z = 4

Question 15.

Measure of angle between the planes 5x – 2y + 3z – 7 = 0 and 15x – 6y + 9z + 5 = 0 is

(A ) 0º

(B ) 30º

(C ) 45º

(D ) 90º

Solution:

(A ) 0º

Question 16.

The direction cosines of the normal to the plane 2x – y + 2z = 3 are

Solution:

(a) \(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\)

Question 17.

The equation of the plane passing through the points (1, -1, 1), (3, 2, 4) and parallel to Y-axis is :

(A ) 3x + 2z – 1 = 0

(B ) 3x – 2z = 1

(C ) 3x + 2z + 1 = 0

(D ) 3x + 2z = 2

Solution:

(B ) 3x – 2z = 1

Question 18.

The equation of the plane in which the line \(\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}\) and \(\frac{x-8}{7}=\frac{y-4}{1}=\frac{z+5}{3}\) lie, is

(A ) 17x – 47y – 24z + 172 = 0

(B ) 17x + 47y – 24z + 172 = 0

(C ) 17x + 47y + 24z +172 = 0

(D ) 17x – 47y + 24z + 172 = 0

Solution:

(A ) 17x – 47y – 24z + 172 = 0

Question 19.

If the line \(\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}\) lies in the plane 3x – 14y + 6z + 49 = 0, then the value of m is:

(A ) 5

(B ) 3

(C ) 2

(D ) -5

Solution:

(A ) 5

Question 20.

The foot of perpendicular drawn from the point (0,0,0) to the plane is (4, -2, -5) then the equation of the plane is

(A ) 4x + y + 5z = 14

(B ) 4x – 2y – 5z = 45

(C ) x – 2y – 5z = 10

(D ) 4x + y + 6z = 11

Solution:

(B ) 4x – 2y – 5z = 45

II. Solve the following :

Question 1.

Find the vector equation of the plane which is at a distance of 5 unit 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 i the length of the perpendicular from origin to the plane, then the vector equation of the plane \(\bar{r} \cdot \hat{n}\) = p

Here, \(\overline{\mathrm{n}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and p = 5

Question 2.

Find the perpendicular distance of the origin from the plane 6x + 2y + 3z – 7 = 0

Solution:

The distance of the point (x_{1}, y_{1}, z_{1}) from the plane ax + by + cz + d 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 6x + 2y + 3z – 7 = 0 is

= 1units.

Question 3.

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.

Solution:

The equation of the plane is 2x + 3y + 6z = 49

Dividing each term by

\(\sqrt{2^{2}+3^{2}+(-6)^{2}}\)

= \(\sqrt{49}\)

= 7

we get

\(\frac{2}{7}\)x + \(\frac{3}{7}\)y – \(\frac{6}{7}\)z = \(\frac{49}{7}\) = 7

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 = \(\frac{3}{7}\), n = \(\frac{6}{7}\)

and length of perpendicular from origin to the plane is p = 7.

the coordinates of the foot of the perpendicular from the origin to the plane are

(lp, ∓, np)i.e.(2, 3, 6)

Question 4.

Reduce the equation \(\bar{r} \cdot(\hat{i}+8 \hat{j}+24 \hat{k})=13\) 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.

Given pane is \(\text { r. }(6 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+24 \hat{\mathrm{k}})=13\) …(1)

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 \(\frac{1}{2}\).

(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 points A(1, -2, 1), B (2, -1, -3) and C (0, 1, 5).

Solution:

The vector equation of the plane passing through three non-collinear points A(\(\bar{a}\)), B(\(\bar{b}\)) and C(\(\bar{c}\)) is \(\bar{r} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) … (1)

Question 6.

Find the Cartesian equation of the plane passing through A(1, -2, 3) and the direction ratios of whose normal are 0, 2, 0.

Solution:

The Cartesian equation of the plane passing through (x_{1}, y_{1}, z_{1}), the direction ratios of whose normal are a, b, c, is

a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0

∴ the cartesian equation of the required plane is

o(x + 1) + 2(y + 2) + 5(z – 3) = 0

i.e. 0 + 2y – 4 + 10z – 15 = 0

i.e. y + 2 = 0.

Question 7.

Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the plane \(\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0\)

Solution:

The cartesian equation of the plane \(\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0\) is 6x + 8y + 7z = 0 The required plane is parallel to it

∴ its cartesian equation is

6x + 8y + 7z = p …(1)

A (7, 8, 6) lies on it and hence satisfies its equation

∴ (6)(7) + (8)(8) + (7)(6) = p

i.e., p = 42 + 64 + 42 = 148.

∴ from (1), the cartesian equation of the required plane is 6x + 8y + 7z = 148.

Question 8.

The foot of the perpendicular drawn from the origin to a plane is M(1, 2,0). 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, 2, 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 \(\bar{m}\) = \(\hat{\mathrm{i}}\).

Normal to the plane is

\(\bar{n}\) = \(\overline{\mathrm{OM}}\) = \(\hat{\mathrm{i}}\)

\(\overline{\mathrm{m}} \cdot \overline{\mathrm{n}}\) = \(\hat{\mathrm{i}}, \hat{i}\) = 5

∴ the vector equation of the required plane is

\(\bar{r} \cdot(\hat{i}+2 \hat{j})\) = 5

Question 9.

A plane makes non zero intercepts a, b, c on the co-ordinates axes. Show that the vector equation of the plane is \(\bar{r} \cdot(b c \hat{i}+c a \hat{j}+a b \hat{k})\) = abc

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

\(\overline{\mathrm{r}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) …(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, , 1)

Question 10.

Find the vector equation of the plane passing through the pointA(-2, 3, 5) and parallel to vectors \(4 \hat{i}+3 \hat{k}\) and \(\hat{i}+\hat{j}\)

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

= (-2)(-4) + (7)(-1) + (5)(4)

= 8 – 7 + 8

= 35

∴ From (1), the vector equation of the required plane is \(\overline{\mathrm{r}} \cdot(-3 \hat{\mathrm{i}}-3 a t \mathrm{j}+4 \hat{\mathrm{k}})\) = 35.

Question 11.

Find the Cartesian equation of the plane \(\bar{r}=\lambda(\hat{i}+\hat{j}-\hat{k})+\mu(\hat{i}+2 \hat{j}+3 \hat{k})\)

Solution:

The equation \(\bar{r}=\bar{a}+\lambda \bar{b}+\mu \bar{c}\) represents a plane passing through a point having position vector \(\overline{\mathrm{a}}\) and parallel to vectors \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\).

Here,

Question 12.

Find the vector equations of planes which pass through A(1, 2, 3), B (3, 2, 1) and make equal intercepts on the co-ordinates axes.

Question is modified

Find the cartesian equations of the planes which pass through A(1, 2, 3), B(3, 2, 1) and make equal intercepts on the coordinate axes.

Solution:

Case 1 : Let all the intercepts be 0.

Then the plane passes through the origin.

Then the cartesian equation of the plane is

ax + by + cz = 0 …..(1)

(1, 2, 3) and (3, 2, 1) lie on the plane.

∴ a + 2b + 3c = 0 and 3a + 2b + c = 0

∴ a, b, c are proportional to 1, -2, 1

∴ from (1), the required cartesian equation is x – 2y + z = 0.

Case 2 : Let the plane make non zero intercept p on each axis.

then its equation is \(\frac{x}{p}+\frac{y}{p}+\frac{z}{p}\) = 1

i.e. x + y + z = p …(2)

Since this plane pass through (1, 2, 3) and (3, 2, 1)

∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p

∴ p = 6

∴ from (2), the required cartesian equation is

x + y + z = 6

Hence, the cartesian equations of required planes are x + y + z = 6 and x – 2y + z = 0.

Question 13.

Find the vector equation of the plane which makes equal non-zero intercepts on the co-ordinates axes and passes through (1, 1, 1).

Solution:

Case 1 : Let all the intercepts be 0.

Then the plane passes through the origin.

Then the vector equation of the plane is ax + by + cz …(1)

(1, 1, 1) lie on the plane.

∴ 1a + 1b + 1c = 0

∴ from (1), the required cartesian equation is x – y + z = 0

Case 2 : Let he plane make non zero intercept p on each axis.

then its equation is \(\frac{\hat{\mathrm{i}}}{p}+\frac{\hat{\mathrm{j}}}{p}+\frac{\hat{\mathrm{k}}}{p}=1\) = 1

i.e. \(\hat{i}+\hat{j}+\hat{k}=p\) = p ….(2)

Since this plane pass through (1, 1, 1)

∴ 1 + 1 + 1 = p

∴ p = 3

∴ from (2), the required cartesian equation is \(\hat{i}+\hat{j}+\hat{k}\) = 3

Hence, the cartesian equations of required planes are \(\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=3\)

Question 14.

Find the angle between planes \(\bar{r} \cdot(-2 \hat{i}+\hat{j}+2 \hat{k})=17\) and \(\bar{r} \cdot(2 \hat{i}+2 \hat{j}+\hat{k})=71\).

Solution:

The acute angle between the planes

= (1)(2) + (1)(1) + (2)(1)

= 2 + 1 + 2

= 5

Also,

Question 15.

Find the acute angle between the line \(\bar{r}=\lambda(\hat{i}-\hat{j}+\hat{k})\) and the plane \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=23\)

Solution:

The acute angle θ between the line \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\) and the plane \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}\) = d is given by

= (2)(2) + (3)(-1) + (-6)(1)

= 4 – 3 – 6

= -5

Question 16.

Show that lines \(\bar{r}=(\hat{i}+4 \hat{j})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\) and \(\bar{r}=(3 \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})\)

Solution:

Question 17.

Find the distance of the point \(3 \hat{i}+3 \hat{j}+\hat{k}\) 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 = \(\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|}\) ……(1)

= (3)(2) + (3)(3) + (1)(-6)

= 6 + 9 – 6

= 9

Question 18.

Find the distance of the point (13, 13, -13) from the plane 3x + 4y – 12z = 0.

Solution:

= 19units.

Question 19.

Find the vector equation of the plane passing through the origln and containing the line \(\bar{r}=(\hat{i}+4 \hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}+\hat{k})\).

Solution:

The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\) … (1)

We can take \(\bar{a}\) = \(\bar{0}\) since the plane passes through the origin.

The point M with position vector \(\bar{m}\) =\(\hat{i}+4 \hat{j}+\hat{k}\) lies on the line and hence it lies on the plane.

.’. \(\overline{\mathrm{OM}}=\bar{m}=\hat{i}+4 \hat{j}+\hat{k}\) lies on the plane.

The plane contains the given line which is parallel to \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)

Let \(\bar{n}\) be normal to the plane. Then \(\bar{n}\) is perpendicular to \(\overline{\mathrm{OM}}\) as well as \(\bar{b}\)

Question 20.

Find the vector equation of the plane which bisects the segment joining A(2, 3, 6) and B( 4, 3, -2) at right angle.

Solution:

The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\) ….(1)

The position vectors \(\bar{a}\) and \(\bar{b}\) of the given points A and B are \(\bar{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}\) and \(\bar{b}=4 \hat{i}+3 \hat{j}-2 \hat{k}\)

If M is the midpoint of segment AB, the position vector \(\bar{m}\) of M is given by

Question 21.

Show thatlines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them.

Solution:

Given lines are x = y, z = 0 and x + y = 0, z = 0.

It is clear that (0, 0, 0) satisfies both the equations.

∴ the lines intersect at O whose position vector is \(\overline{0}\)

Since z = 0 for both the lines, both the lines lie in XY- plane.

Hence, we have to find equation of XY-plane.

Z-axis is perpendicular to XY-plane.

∴ normal to XY plane is \(\hat{k}\).

0(\(\overline{0}\)) lies on the plane.

By using \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\), the vector equation of the required plane is \(\bar{r} \cdot \hat{k}=\overline{0} \cdot \bar{k}\)

i.e. \(\bar{r} \cdot \hat{k}=0\).

Hence, the given lines intersect each other and the vector equation of the plane determine by them is \(\bar{r} \cdot \hat{k}=0\).