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# Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.10

• Last Updated : 03 Nov, 2022

### 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 My Personal Notes arrow_drop_up