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