Class 12 RD Sharma Solutions - Chapter 29 The Plane - Exercise 29.10

Question 1. Find the distance between the parallel planes 2x - y + 3z - 4 = 0 and 6x - 3y + 9z + 13 = 0

Solution:

Let P(x₁, y₁, z₁) be any point on plane 2x – y + 3z – 4 = 0.

⟹ 2x₁ – y₁ + 3z₁ = 4 (equation-1)

Distance between (x₁, y₁, z₁) and the plane

6x – 3y + 9z + 13 = 0:

As we know that, the distance of point (x₁, y₁, z₁) from the plane π: ax + by + cz + d = 0 is given by:

p = 

Now, substitute the values, we get

p = 

= 

=   [by using equation 1]

= 

Therefore, the distance between the parallel planes 2x - y + 3z - 4 = 0 and 6x - 3y + 9z + 13 = 0 is  units.

Question 2. Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane 2x - 3y + 5z + 7 = 0. Also find the distance between the two planes.

Solution:

Since the plane is parallel to 2x – 3y + 5z + 7 = 0, it must be of the form:

2x – 3y + 5z + θ = 0

It is given that,

The plane passes through (3, 4, –1)

⟹ 2(3) – 3(4) +5(–1) + θ = 0

θ = -11

Thus, 

The equation of the plane is as follows:

2x – 3y + 5z – 11 = 0

Distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1):

As we know that, the distance of point (x₁, y₁, z₁) from the plane π: ax + by + cz + d = 0 is given by:

p = 

Now, after substituting the values, we will get

= 

= 

Therefore, the distance of the plane 2x - 3y + 5z + 7 = 0 from (3, 4, -1) is 

Question 3. Find the equation of the plane mid-parallel to the planes 2x - 2y + z + 3 = 0 and 2x - 2y + z + 9 = 0

Solution:

Given:

Equation of planes: 

π₁= 2x – 2y + z + 3 = 0

π₂= 2x – 2y + z + 9 = 0

Let the equation of the plane mid–parallel to these planes be:

π₃: 2x – 2y + z + θ = 0

Now,

Let P(x₁, y₁, z₁) be any point on this plane,

⟹ 2(x₁) – 2(y₁) + (z₁) + θ = 0 ---(equation-1)

As we know that, the distance of point (x₁, y₁, z₁) from the plane π: ax + by + cz + d = 0 is given by:

p = 

Distance of P from π₁:

p = 

=    (By using equation 1)

Similarly,

DIstance of q from π₂:

q = 

=    (By using equation 1)

As π₃ is mid-parallel is π₁ and π₂:

p = q 

So,

Now square on both sides, we get

(3 - θ)² = (9 - θ)²

9 - 2×3×θ + θ² = 81 - 2×9×θ + θ²

θ = 6

Now, substitute the value of θ = 6 in equation 2x - 2y + z + θ = 0, we get

Hence, the equation of the mid-parallel plane is 2x - 2y + z + 6 = 0

Question 4. Find the distance between the planes  and 

Solution:

Let  be the position vector of any point P on the plane

So, 

 ---(equation 1)

As we know that, the distance of from  the plane  is given by:

p = 

Length of perpendicular from is given by substituting the values of, we get

p = 

= 

= 

p = 

Therefore, the distance between the planes

 and  is 

Summary

Exercise 29.10 in RD Sharma's Class 12 Chapter 29 on The Plane likely focuses on finding the equation of a plane under various conditions. This exercise probably covers deriving plane equations when given information such as a point and a normal vector, three non-collinear points, the intercepts on coordinate axes, or when the plane is parallel or perpendicular to another plane or line. Students are expected to apply vector and coordinate geometry concepts to formulate these equations in different scenarios.