RD Sharma Class 12 Ex 29.2 Solutions Chapter 29 The Plane

Here we provide RD Sharma Class 12 Ex 29.2 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 29.2 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.2
CategoryRD Sharma Solutions

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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