# 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 â€“ y + 3z â€“ 4 = 0.âŸ¹ 2x

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

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

As we know that, the distance of point (x

_{1}, y_{1}, z_{1}) from the plane Ï€: 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 â€“ 3y + 5z + 7 = 0, it must be of the form:

2x â€“ 3y + 5z + Î¸ = 0

It is given that,

The plane passes through (3, 4, â€“1)

âŸ¹ 2(3) â€“ 3(4) +5(â€“1) + Î¸ = 0

Î¸ = -11

Thus,

The equation of the plane is as follows:

2x â€“ 3y + 5z â€“ 11 = 0

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

As we know that, the distance of point (x

_{1}, y_{1}, z_{1}) from the plane Ï€: 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:

Ï€

_{1}= 2x â€“ 2y + z + 3 = 0Ï€

_{2}= 2x â€“ 2y + z + 9 = 0Let the equation of the plane midâ€“parallel to these planes be:

Ï€

_{3}: 2x â€“ 2y + z + Î¸ = 0Now,

Let P(x

_{1}, y_{1}, z_{1}) be any point on this plane,âŸ¹ 2(x

_{1}) â€“ 2(y_{1}) + (z_{1}) + Î¸ = 0 —(equation-1)As we know that, the distance of point (x

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

Distance of P from Ï€

_{1}:p =

= (By using equation 1)

Similarly,

DIstance of q from Ï€

_{2}:q =

= (By using equation 1)

As Ï€

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

So,

Now square on both sides, we get

(3 – Î¸)

^{2}= (9 – Î¸)^{2}9 – 2Ã—3Ã—Î¸ + Î¸

^{2}= 81 – 2Ã—9Ã—Î¸ + Î¸^{2}Î¸ = 6

Now, substitute the value of Î¸ = 6 in equation 2x – 2y + z + Î¸ = 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

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