# 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(x

_{1}, y_{1}, z_{1}) be any point on plane 2x – y + 3z – 4 = 0.⟹ 2x

_{1}– y_{1}+ 3z_{1}= 4 (equation-1)Distance between (x

_{1}, y_{1}, z_{1}) and the plane6x – 3y + 9z + 13 = 0:

As we know that, the distance of point (x

_{1}, y_{1}, z_{1}) 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 (x

_{1}, y_{1}, z_{1}) 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:

π

_{1}= 2x – 2y + z + 3 = 0π

_{2}= 2x – 2y + z + 9 = 0Let the equation of the plane mid–parallel to these planes be:

π

_{3}: 2x – 2y + z + θ = 0Now,

Let P(x

_{1}, y_{1}, z_{1}) be any point on this plane,⟹ 2(x

_{1}) – 2(y_{1}) + (z_{1}) + θ = 0 —(equation-1)As we know that, the distance of point (x

_{1}, y_{1}, z_{1}) from the plane π: ax + by + cz + d = 0 is given by:p =

Distance of P from π

_{1}:p =

= (By using equation 1)

Similarly,

DIstance of q from π

_{2}:q =

= (By using equation 1)

As π

_{3}is mid-parallel is π_{1}and π_{2}:p = q

So,

Now square on both sides, we get

(3 – θ)

^{2}= (9 – θ)^{2}9 – 2×3×θ + θ

^{2}= 81 – 2×9×θ + θ^{2}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 pont 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