NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.3

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

Question 1.
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2
(b) x + y + z = 1
(c) 2x + 3y – z = 5
(d) 5y + 8 = 0
Solution:
a. The equation can be written as 0x + 0y + 1z = 2,
which is in the normal form Hence the direction cosines of the normal to the plane are 0, 0, 1 and distance from the orgin = 2

b. Equation of the plane is x + y + z = 1 … (1)
The direction ratios of the normal to the plane are 1, 1, 1
∴ Direction cosines are
\(\frac{1}{\sqrt{1+1+1}}, \frac{1}{\sqrt{1+1+1}}, \frac{1}{\sqrt{1+1+1}}\)
\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
Dividing (1) by \(\sqrt{3}\), we get
\(\frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}}=\frac{1}{\sqrt{3}}\)
This is of the form lx + my + nx = d where d is the distance from the orgin.
∴ Distance from the orgin = \(\frac{1}{\sqrt{3}}\)

c. The direction ratios of the normal to the plane are 2, 3, -1
The direction cosines are
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 1

d. The direction ratios of the normal to the plane are 0, 5, 0
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 2

Question 2.
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \(3\hat { i } +5\hat { j } -6\hat { k }\)
Solution:
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 3

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

Question 3.
Find the Cartesian equation of the following planes.
(a) \(\overrightarrow { r } \cdot (\hat { i } +\hat { j } -\hat { k) }\) = 2
(b) \(\overrightarrow { r } \cdot (\hat { 2i } +3\hat { j } -4\hat { k) }\) = 1
(c) \(\overrightarrow { r } \cdot [(s-2t)\hat { i } +(3-t)\hat { j } +(2s+t)\hat { k) }\) = 15
Solution:
Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) be the position vector of any point on the plane.
\(\overrightarrow { r } \cdot (\hat { i } +\hat { j } -\hat { k) }\) = 2
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})\)
x + y – z = 2 is the cartesian equation.

b. Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ \(\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})\) = 1,
\((x \hat{i}+\hat{y j}+z \hat{k}) \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})\) = 1
2x + 3y – 4z = 1 is the cartesian equation.

c. Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
\(\vec{r} \cdot((s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k})\) = 15
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot((s-2 t) \hat{i}+(3-t) \hat{j}\) + \((2 s+t) \hat{k})\) = 15
(s – 2t)x + (3 – t)y + (2s + t)z = 15, is the cartesian equation.

Question 4.
In the following cases find the coordinates of the foot of perpendicular drawn from the origin
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Solution:
Let A be the foot of the perpendicular drawn from the orgin O.
Equation of the plane is 2x + 3y + 4z – 12 = 0
∴ The d.r’s of the normal to the plane are 2, 3, 4
Equation of the line OA is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
\(\frac{x-0}{2}=\frac{y-0}{3}=\frac{z-0}{4}\)
i.e. \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\lambda\)
∴ Any point on the line OA is (2λ, 3λ, 4λ) Let it be A.
A satisfies the equation 2x + 3y + 4z – 12 = 0
4λ + 9λ + 16λ – 12 = 0
29λ – 12 = 0
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 4

Another Method:
The equation of the plane is 2x + 3y + 42 = 12 … (1)
Dividing (1) by \(\), we get
\(\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29}\) is of the form lx + my + nz = d
The direction cosines of the normal to the plane are l = \(\frac{2}{\sqrt{29}}\), m = \(\frac{3}{\sqrt{29}}\), n = \(\frac{4}{\sqrt{29}}\)
and the distance from the origin to the plane d = \(\frac{12}{\sqrt{29}}\)
Foot of the perpendicular = (Id, md, nd)
= \(\left(\frac{2}{\sqrt{29}}\left(\frac{12}{\sqrt{29}}\right), \frac{3}{\sqrt{29}}\left(\frac{12}{\sqrt{29}}\right), \frac{4}{\sqrt{29}}\left(\frac{12}{\sqrt{29}}\right)\right)\)
= \(\left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right)\)

b. The equation of the plane is , 3y + 4z = 6 … (1)
Dividing (1) by \(\sqrt{3^{2}+4^{2}}\) = 5, we get
\(\frac{3}{5} y+\frac{4}{5} z=\frac{6}{5}\), is in the form lx + my + nz = d
∴ l = 0, m = \(\frac { 3 }{ 5 }\), n = \(\frac { 4 }{ 5 }\), d = \(\frac { 6 }{ 5 }\)
The foot of the perpendicular from the origin = (Id, md, nd)
= \(\left(0\left(\frac{6}{5}\right), \frac{3}{5}\left(\frac{6}{5}\right), \frac{4}{5}\left(\frac{6}{5}\right)\right)\)
= \(\left(0, \frac{18}{25}, \frac{24}{25}\right)\)

c. The equation of the plane is
x + y + z = 1 … (1)
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 5

d. Equation of the plane is
5y = – 8 or – 5y = 8
Dividing (1) by \(\sqrt{(-5)^{2}}\) = 5, we get
\(\frac{-5}{5} y=\frac{8}{5} \quad \text { or } \quad-y=\frac{8}{5}\), is in the form
lx + my + nz = d
l = 0, m = – 1, n = 0, d = \(\frac { 8 }{ 5 }\)
The foot of the perpendicular from the origin = (Id, md, nd)
= \(\left(0\left(\frac{8}{5}\right),-1\left(\frac{8}{5}\right), 0\left(\frac{8}{5}\right)\right)\)
= \(\left(0, \frac{-8}{5}, 0\right)\)

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

Question 5.
Find the vector and cartesian equation of the planes
(a) that passes through the point (1,0, -2) and the normal to the plane is \(\hat { i } +\hat { j } -\hat { k }\)
(b) that passes through the point (1,4,6) and the normal vector to the plane is \(\hat { i } -2\hat { j } +\hat { k }\)
Solution:
(a) The plane passes through the point (1, 0, – 2)
∴ \(\vec{a}=\hat{i}+0 \hat{j}-2 \hat{k}\)
The normal vector to the plane \(\vec{n}=\hat{i}+\hat{j}-\hat{k}\)
The vector equation of the plane passing through \(\vec{a}\) and perpendicular to \(\vec{n}\) is (r – a).n = 0
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 6
The cartesian equation is x + y – z – 3 = 0

(b) The plane passes through (1, 4, 6) and the direction ratios of the normal to the plane are 1, -2, 1.
The cartesian equation of the plane passing through (x1, T1, Z1) and perpendicular to the line with direction ratios a, b, c is
a(x – x1) + b(y – y1) + c(z – z1) = 0
Hence the cartesian equation of the plane is 1 (x – 1) – 2(y – 4) + 1(z – 6) = 0
x – 2y + z + 1 = 0
The vector equation of the plane is
\(\vec{r} \cdot(\hat{i}-2 \hat{j}+\hat{k})+1=0\)

Question 6.
Find the equations of the planes that passes through three points
(a) (1, 1, – 1) (6, 4, – 5), (- 4, -2, 3)
(b) (1, 1, 0), (1, 2, 1), (-2, 2, -1)
Solution:
Let the equation of plane passing through (1, 1, – 1) be
a(x – 1) + b(y – 1) + c(z + 1 ) = 0 … (1)
Since (6, 4, – 5) is a point on (1), we get
a(6 – 1) + b(4 – 1) + c(- 5 + 1) = 0
i.e., 5a + 3b – 4c = 0 … (2)
Since (- 4, – 2, 3) is a point on (1), we get
a(- 4 – 1) + b(- 2 – 1) + c(3 + 1) = 0
– 5a – 3b + 4c = 0
5a + 3b – 4c = 0 … (3)
From (2) and (3), by the rule of cross multiplication, we get
\(\frac{a}{-12+12}=\frac{b}{-20+20}=\frac{c}{15-15}\)
\(\frac{a}{0}=\frac{b}{0}=\frac{c}{0}\)
a = 0, b = 0 and c = 0
Hence, there is no unique plane passing through the given points.

Another Method:
The plane passes through
(1, 1, – 1),(6, 4, – 5), (- 4, – 2, 3)
The equation of the plane passing through
\(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right) \text { and }\left(x_{3}, y_{3}, z_{3}\right)\) is
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 7
R2 and R3 are proportional.
Hence we get 0 = 0
So we cannot find the equation of the plane passing through these points in a unique way.

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

Question 7.
Find the intercepts cut off by the plane 2x+y-z = 5.
Solution:
Equation of the plane is 2x + y- z = 5 x y z
Dividing by 5:
\(\Rightarrow \frac { x }{ \frac { 5 }{ 2 } } +\frac { y }{ 5 } -\frac { z }{ -5 }\) = 1
∴ The intercepts on the axes OX, OY, OZ are \(\frac { 5 }{ 2 }\), 5, – 5 respectively.

Question 8.
Find the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane.
Solution:
Any plane parallel to ZOX plane is y=b where b is the intercept on y-axis.
∴ b = 3.
Hence equation of the required plane is y = 3.

Question 9.
Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2,2,1).
Solution:
Given planes are:
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0
Any plane through their intersection is
3x – y + 2z – 4 + λ(x + y + z – 2) = 0
point (2, 2, 1) lies on it,
∴ 3 x 2 – 2 + 2 x 1 – 4 + λ(2 + 2 + 1 – 2) = 0
⇒ λ = \(\frac { -2 }{ 3 }\)
Now required equation is 7x – 5y + 4z – 8 = 0

Question 10.
Find the vector equation of the plane passing through the intersection of the planes \(\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =7,\overrightarrow { r } \cdot \left( 2\hat { i } +5\hat { j } +3\hat { k } \right) =9\) and through the point (2, 1, 3).
Solution:
Here \(\vec{n}_{1}=2 \hat{i}+2 \hat{j}-3 \hat{k}, d_{1}=7\)
\(\vec{n}_{2}=2 \hat{i}+5 \hat{j}+3 \hat{k}, d_{2}=9\)
The required vector equation is
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 8
is the required vector equation of the plane.

Another method:
The equation of the planes are
2x + 2y – 3z = 7 … (1)
2x + 5y + 3z = 9 … (2)
The equation of the plane through the inter-section of (1) and (2) is (2x + 2y – 3z) + λ(2x + 5y + 3z) = 7 + 9λ … (3)
Since the plane (3) passes through (2, 1, 3), we get
(2(2) + 2(1) – 3(3)) + λ (2(2) + 5(1) + 3(3)) = 7 + 9λ
– 3 + 18λ = 7 + 9λ
9 λ = 10 ∴ λ = \(\frac { 10 }{ 9 }\)
(3) gives the cartesian equation of the plane.
(3) → (2x + 2y – 3z) + \(\frac { 10 }{ 9 }\) (2x + 5y + 3z)
= 7 + 9(\(\frac { 10 }{ 9 }\))
Multiplying by 9, we get 18x + 18y – 27z + 20x + 50y + 30z = 63 + 90 38x + 68y + 3z = 153
The vector equation is
\(\bar{r} \cdot(38 \hat{i}+68 \hat{j}+3 \hat{k})\) = 153

Question 11.
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = o.
Solution:
The equation of the required plane is (x + y + z – 1) + λ(2x + 3y + 4z – 5) = 0
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 9

Question 12.
Find the angle between the planes whose vector equations are \(\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =5,\overrightarrow { r } \cdot \left( 3\hat { i } -3\hat { j } +5\hat { k } \right)\) = 3
Solution:
The angle θ between the given planes is
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 10

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

Question 13.
In the following determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angle between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0.
Solution:
a. Comparing with the general equation, we get
a1 = 7 b1 = 5 c1 = 6 and
a2 = 3 b2 = – 1 c2 = – 10
a1a2 + b1b2 + c1c2 = 7(3) + 5(- 1) + 6(- 10) ≠ 0
∴ The lines are not perpendicular
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 11

b. a1 = 2 b1 = 1 c1 = 3 and
a2 = 1 b2 = – 2 c2 = 0
a1a2 + b1b2 + c1c2 = 2 – 2 + 0 = 0
∴ The two lines are perpendicular.

c. a1 = 2 b1 = – 2 c1 = 4 and
a2 = 3 b2 = – 3 c2 = 6
\(\frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{2}{3}, \frac{c_{1}}{c_{2}}=\frac{4}{6}=\frac{2}{3}\)
i.e., \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
∴ The two lines are parallel.

d. a1 = 2 b1 = – 1 c1 = 3 and
a2 = 2 b2 = – 1 c2 = 3
\(\frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=1, \frac{c_{1}}{c_{2}}=1\)
∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
∴ The two lines are parallel.

e. a1 = 4 b1 = 8 c1 = 1 and
a2 = 0 b2 = 1 c2 = 1
a1a2 + b1b2 + c1c2 = 2(3) + – 2(3) + 4(6) ≠ 0
∴ The two lines are perpendicular.
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 12

Question 14.
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point           Plane
(a) (0, 0,0) 3x – 4y + 12z = 3
(b) (3,-2,1) 2x – y + 2z + 3 = 0.
(c) (2,3,-5) x + 2y – 2z = 9
(d) (-6,0,0) 2x – 3y + 6z – 2 = 0
Solution:
(a) Given plane: 3x – 4y + 12z – 3 = 0
∴ The distance from the point (0, 0, 0) to
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 13

b. The equation of the plane is
2x – y + 2z + 3 = 0
∴ The distance from the point (3, – 2, 1) to the given plane
= \(\left|\frac{2(3)-1(-2)+2(1)+3}{\sqrt{4+1+4}}\right|=\frac{13}{3}\)

c. The equation of the plane is x + 2y – 2z – 9 = 0
a = 1, b = 2, c = – 2, d = 9
∴ The distance from the point (2, 3, – 5) to the plane
= \(\left|\frac{a x_{1}+b y_{1}+c z_{1}-d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|=\left|\frac{2+2(3)-2(-5)-9}{\sqrt{1+4+4}}\right|\)
= \(\frac{9}{3}\) = 3

d. The equation of the plane is 2x – 3y + 6z – 2 = 0
∴ The perpendicular distance from the point (- 6, 0, 0) to the plane
= \(\left|\frac{2(-6)-3(0)+6(0)-2}{\sqrt{4+9+36}}\right|\)
= \(\left|\frac{14}{\sqrt{49}}\right|=\frac{14}{7}\) = 2

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