NCERT Solutions for Maths Class 12 Exercise 11.3

# NCERT Solutions for Maths Class 12 Exercise 11.3

## NCERT Solutions for Maths Class 12 Exercise 11.3

Hello Students. Welcome to maths-formula.com. In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 11.3.

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Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

NCERT Solutions for Maths Class 12 Exercise 11.3 helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

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NCERT Solutions for Maths Class 12 Exercise 11.3 are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

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NCERT Solutions for Maths Class 12 Exercise 11.1

## NCERT Solutions for Maths Class 12 Exercise 11.3

Maths Class 12 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) Direction ratios of the normal to the plane are 0, 0, 1.
Here, a = 0, b = 0, c = 1

Maths Class 12 Ex 11.3 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 3i + 5j – 6k.

Solution:

Maths Class 12 Ex 11.3 Question 3.

Find the Cartesian equation of the following planes.

Solution:

(a) Let r is the position vector of any arbitrary point P(x, y, z) on the plane.

Maths Class 12 Ex 11.3 Question 4.

In the following cases, find the coordinates of the foot of the 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)

Maths Class 12 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 i + jk.
(b) that passes through the point (1, 4, 6) and the normal vector to the plane is i – 2j + k.

Solution:

(a) Normal to the plane is i + jk and passes through (1, 0, –2)

Maths Class 12 Ex 11.3 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

Maths Class 12 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
Dividing by 5:

The intercepts on the axes OX, OY, OZ are 5/2, 5, –5, respectively.

Maths Class 12 Ex 11.3 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.

Maths Class 12 Ex 11.3 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 + Î»(x + y + z – 2) = 0    … (i)
Point (2, 2, 1) lies on it,
3 × 2 – 2 + 2 × 1 – 4 + Î»(2 + 2 + 1 – 2) = 0
Thus, Î» = –2/3
Putting the value of Î» in equation (i), we get the required equation is 7x – 5y + 4z – 8 = 0.

Maths Class 12 Ex 11.3 Question 10.

Find the vector equation of the plane passing through the intersection of the planes r.(2i + 2j – 3k) = 7 and r.(2i + 5j + 3k) = 9 and through the point (2, 1, 3).

Solution:

The equation of the plane passing through the line of intersection of the planes

Maths Class 12 Ex 11.3 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)

Maths Class 12 Ex 11.3 Question 12.

Find the angle between the planes whose vector equations are r.(2i + 2j – 3k) = 5 and r.(3i – 3j + 5k) = 3.

Solution:

The angle Î¸ between the given planes is

Maths Class 12 Ex 11.3 Question 13.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles 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)

Maths Class 12 Ex 11.3 Question 14.

In the following cases, find the distance of each of the given points from the corresponding given plane.
(a) Point (0, 0, 0); Plane 3x – 4y + 12z = 3
(b) Point (3, –2, 1); Plane 2x – y + 2z + 3 = 0
(c) Point (2, 3, –5); Plane x + 2y – 2z = 9
(d) Point (–6, 0, 0); Plane 2x – 3y + 6z – 2 = 0

Solution:

(a) Given plane: 3x – 4y + 12z – 3 = 0