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

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3.

Board CBSE
Textbook NCERT
Class Class 12
Subject Maths
Chapter Chapter 11
Chapter Name Three Dimensional Geometry
Exercise Ex 11.3
Number of Questions Solved 14
Category NCERT Solutions

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

Ex 11.3 Class 12 Maths 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) Direction ratios of the normal to the plane are 0,0,1
=> a = 0, b = 0, c = 1

Ex 11.3 Class 12 Maths 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:

Ex 11.3 Class 12 Maths 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:
(a) \overrightarrow { r } is the position vector of any arbitrary point P (x, y, z) on the plane.

Ex 11.3 Class 12 Maths 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:
(a) Let N (x1, y1, z1) be the foot of the perpendicular from the origin to the plane 2x+3y+4z-12 = 0
∴ Direction ratios of the normal are 2, 3, 4.
Also the direction ratios of ON are (x1,y1,z1)


Ex 11.3 Class 12 Maths 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) Normal to the plane is i + j – k and passes through (1,0,-2)

Ex 11.3 Class 12 Maths 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:
(a) The plane passes through the points (1,1,-1) (6,4,-5), (-4,-2,3)
Let the equation of the plane passing through(1,1,-1)be

Ex 11.3 Class 12 Maths 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

Ex 11.3 Class 12 Maths 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.

Ex 11.3 Class 12 Maths 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

Ex 11.3 Class 12 Maths 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:
Equation of the plane passing through the line of intersection of the planes

Ex 11.3 Class 12 Maths 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 = 0.
Solution:
Given planes are
x + y + z – 1 = 0 …(i)
2x + 3y + 4z – 5 = 0 …(ii)
x – y + z = 0 ….(iii)

Ex 11.3 Class 12 Maths 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

Ex 11.3 Class 12 Maths 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) Direction ratios of the normal of the planes 7x + 5y + 6z + 30 = 0 are 7,5,6
Direction ratios of the normal of the plane 3x – y – 10z + 4 = 0 are 3,-1,-10
The plane 7x + 5y + 6z + 30 = 0 …(i)
3x – y – 10z + y = 0 …(ii)


Ex 11.3 Class 12 Maths 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

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