RD Sharma Class 12 Ex 29.2 Solutions Chapter 29 The Plane

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RD Sharma Class 12 Ex 29.2 Solutions Chapter 29 The Plane

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Write the equation of the plane whose intercepts on the coordinate axes are 2, -3 and 4.

Solution:

Given: Intercept on the coordinate axes are 2,-3 and 4

We represent equation of a plane whose intercepts on 

the coordinate axes are p, q and r respectively as follows,

(x / p) + (y / q) + (z / r) = 1          -Equation (1)

Here, p = 2, q = -3, r = 4,

The required equation of plane is 

x / 2 + y / -3 + z / 4  = 1 

6x – 4y + 3z / 12 = 1 

6x – 4y + 3z = 12

Question 2(i). Reduce the equation of the following planes to intercept form and find the intercepts on the coordinate axes: 4x + 3y – 6z -12 = 0.

Solution:

Reduce the equation 4x + 3y – 6z -12 = 0 to intercept form

 4x + 3y – 6z -12 = 0         – Equation (1)

Divide equation(1) by 12

4x / 12 + 3y / 12 – 6z / 12 = 12 / 12 

x / 3 + y / 4 + z / -2 = 1          -Equation (2)

This is in the form x / a + y / b + z / c = 1          -Equation (3)

On comparing equation (2) and equation (3), we get

a = 3, b = 4, c = -2

So, the intercepts on coordinate axes are 3, 4, -2.

Question 2(ii). Reduce the equation of the following planes to intercept form and find the intercepts on the coordinate axes: 2x + 3y – z = 6. 

Solution: 

Reduce 2x + 3y – z = 6 in intercept form :

2x + 3y – z = 6         -Equation(1)

Divide equation (1) by 6,

2x / 6 + 3y / 6 – z / 6 = 6 / 6

x / 3 + y / 2 + z / -6 = 1          -Equation (2)

Equation of intercept form of plane with a, b, c as intercepts on coordinate axes is,

x / a + y / b + z / c = 1          -Equation (3)

On comparing equation (2) and equation (3)

We get a = 3, b = 2, c = -6

So, The intercepts on coordinate axes by given plane are 3, 2, -6

Question 2(iii). Reduce the equation of the following planes to intercept form and find the intercepts on the coordinate axes: 2x – y + z = 5.

Solution: 

To find intercepts on coordinate axes by plane 2x – y + z = 5

2x – y + z = 5         -Equation(1)

Divide equation (1) by 5,

2x / 5 – y / 5 + z / 5 = 5 / 5

x / ( 5 /2 ) + y /-5 + z / 5 = 1          -Equation (2)

Equation of intercept form of plane with a, b, c as intercepts on coordinate axes is,

x / a + y / b + z / c = 1          -Equation (3)

On comparing equation (2) and equation (3)

We get a = 5 / 2, b = -5, c = 5

So, the intercepts on coordinate axes by given plane as 5 / 2, -5, 5.

Question 3. Find the equation of a plane which meets the axes at A, B, and C given that the centroid of the triangle ABC is the point (α, β, γ).

Solution: 

Given that plane meets axes in A, B, and C

Let, A = (a, 0, 0), B = (0, b, 0), c = (0, 0, c)

We have centroid of ABC as (α, β, γ). Centroid of plane ABC is given by,

Centroid = (x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3, (z1 + z2 + z3) / 3

(α, β, γ) = [(a + 0 + 0) /3, (0 + b + 0) / 3, (0 + 0 + c) / 3]

(α, β, γ) = [a / 3, b / 3, c / 3]

We get,

 a / 3 = α ⇒     a = 3α         -Equation(1)

b / 3 = β ⇒     b = 3β         -Equation(2)

c / 3 = γ ⇒     c = 3γ         -Equation(3)

Equation of intercept form of plane with a, b, c 

as intercepts on coordinate axes is,

x / a + y / b + z / c = 1  

Put a, b, c from equation (1),(2) and (3),

x / 3α + y / 3β + z / 3γ = 1

Multiplying by 3 on both sides,

3x / 3α + 3y / 3β + 3z / 3γ = 3

x / α + y / β + z / γ = 3

Question 4. Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinates axes.

Solution: 

Intercepts on coordinate axes are equal,

Equation of intercept form of plane with a, b, c as intercepts on coordinate axes is,

x / a + y / b + z / c = 1  

From given condition let, a = b = c = p (say)

x / p + y / p + z / p = 1

x + y + z / p = 1

x + y + z = p         -Equation(1)

Plane is passing through the point (2, 4, 6). By using equation(1)

x + y + z = p

2 + 4 + 6 = p

12 = p

Substitute p in equation(1) 

x + y + z = 12

So the required equation of plane is given by,

x + y + z = 12

Question 5. A plane meets the coordinate axes at A, B, and C respectively, such that the centroid of the triangle ABC is (1, -2, 3). Find the equation of the plane.

Solution: 

Given that plane meets the coordinate axes at A, B, C.Centroid of plane ABC is (1, -2, 3)

Equation of intercept form of plane with a, b, c as intercepts on coordinate axes is,

x / a + y / b + z / c = 1         -Equation(1)

Centroid =(x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3, (z1 + z2 + z3) / 3

(1, -2, 3) = [(a + 0 + 0) /3, (0 + b + 0) / 3, (0 + 0 + c) / 3]

(1, -2, 3) = [a / 3, b / 3, c / 3]

Now by comparing we get,

 a / 3 = 1 ⇒    a = 3         -Equation(2)

b / 3 = -2 ⇒    b = -6         -Equation(3)

c / 3 = 3 ⇒   c = 9         -Equation(4)

Substitute a, b, c in equation (1) to get the equation of required plane

x / 3 + y / -6 + z / 9 = 1

6x – 3x + 2z / 18 = 1

6x – 3x + 2z = 18

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