# RD Sharma Class 12 Ex 29.5 Solutions Chapter 29 The Plane

Here we provide RD Sharma Class 12 Ex 29.5 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.5 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 29.5 Solutions Chapter 29 The Plane

### Question 1. Find the vector equation of the plane passing through the points (1, 1, 1), (1, -1, 1) and (-7, -3, -5)

Solution:

Given that, plane is passing through

(1, 1, 1), (1, -1, 1) and (-7, -3, -5)

We know that, equation of plane passing through 3 points,

(x – 1)(12 – 0) – (y – 1)(0 – 0) + (z – 1)(0 – 16) = 0

(x – 1)(12) – (y – 1)(0) + (z – 1)(-16) = 0

12x – 12 – 0 – 16z + 16 = 0

12x – 16z + 4 = 0

Dividing by 4,

3x – 4z + 1 = 0

Equation of the required plane,

### Question 2. Find the vector equation of the plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3).

Solution:

Let  P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3) be the three points on a plane having position vectors respectively. Then the vectors and are in the same plane. Therefore, is a vector perpendicular to the plane.

Let  = Similarly,

Thus

The plane passes through the point P with position vector Thus, its vector equation is

### Question 3. Find the vector equation of the plane passing through the points A(a, 0, 0), B(0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from origin, prove that Solution:

Let A(a, 0, 0), B(0, b, 0) and C(0, 0, c) be three points on a plane having their position vectors respectively. Then vectors and are in the same plane. Therefore, is a vector perpendicular to the plane.

Let Similarly,

Thus

= |  -a  b  0 |

-a  0  c

The plane passes through the point P with position vector Thus, the vector equation in the normal form is

The vector equation of a plane normal to the unit vector and at a distance ‘d’ from the origin is ….(2).

Given that the plane is at a distance ‘p’ from the origin.

Comparing equations (1) and (2), we have,

d = p = ### Question 4. Find the vector equation of the plane passing through the points (1, 1, -1), (6, 4, -5) and (-4, -2, 3).

Solution:

Let P(1, 1, -1), Q(6, 4, -5) and R(-4, -2, 3) be three points on a plane having position vectors respectively. Then the vectors are in the same plane. Therefore, is a vector perpendicular to the plane.

Let Similarly,

Thus

Here, Therefore, the given points are collinear.

Thus, where, 5a + 3b – 4c = 0

The plane passes through the point P with position vector Thus, its vector equation is , where, 5a + 3b – 4c = 0

### Question 5. Find the vector equation of the plane passing through the points

Solution:

Let A, B, C be the points with position vector respectively. Then = Position vector of B – Position vector of A = Position vector of C – Position vector of B

A vector normal to A, B, C is a vector perpendicular to As we know that, equation of a plane passing through vector and perpendicular to vector is given by,

Put and in equation (1)

= (3)(-36) + (4)(-8) + (2)(28)

= -108 – 32 + 56

= -140 + 56 = -84

Dividing by (-4), we will get

Equation of required plane is,

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