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 = $\left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|$
Now, substitute the values, we get
p = $\left|\frac{(6)(x_1)+(-3)(y_1)+(9)(z_1)+13}{\sqrt{6^2+(-3)^2+9^2}}\right|$
= $\left|\frac{3(2x_1-y_1+3z_1)+13}{\sqrt{6^2+(-3)^2+9^2}}\right|$
= $\left|\frac{3(4)+13}{3\sqrt{14}}\right|$ [by using equation 1]
= $\frac{25}{3\sqrt{14}}$
Therefore, the distance between the parallel planes 2x – y + 3z – 4 = 0 and 6x – 3y + 9z + 13 = 0 is $\frac{25}{3\sqrt{14}}$ 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 = $\left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|$
Now, after substituting the values, we will get
= $\left|\frac{(2)(3)+(-3)(4)+(5)(-1)+7}{\sqrt{2^2+(-3)^2+5^2}}\right|$
= $\frac{4}{\sqrt{38}}$
Therefore, the distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, -1) is $\frac{4}{\sqrt{38}}$

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 = $\left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|$
Distance of P from π₁:
p = $\left|\frac{(2x_1-2y_2+z_1)+3}{\sqrt{2^2+(-2)^2+1^2}}\right|$
= $\left|\frac{(-θ)+3}{3}\right|$ (By using equation 1)
Similarly,
DIstance of q from π₂:
q = $\left|\frac{(2x_1-2y_2+z_1)+9}{\sqrt{2^2+(-2)^2+1^2}}\right|$
= $\left|\frac{(-θ)+9}{3}\right|$ (By using equation 1)
As π₃ is mid-parallel is π₁ and π₂:
p = q
So,
$\left|\frac{(-θ)+3}{3}\right|=\left|\frac{-θ +9}{3}\right|$
Now square on both sides, we get
$\left(\frac{(-θ)+3}{3}\right)^2=\left(\frac{-θ+9}{3}\right)^2$
(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 $\vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0$ and $\vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0$

Solution:

Let $\vec{a}$ be the position vector of any point P on the plane
$\vec{r}(\hat{i}+2\hat{j}+3\hat{k})+7=0$
So,
$\vec{a}(\hat{i}+2\hat{j}+3\hat{k})+7=0$ —(equation 1)
As we know that, the distance of from $\vec{a}$ the plane $\vec{r}.\vec{n}-d=0$ is given by:
p = $\left|\frac{\vec{a}.\vec{n}-d}{|n|}\right|$
Length of perpendicular from $P(\vec{a}) to plane \vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0$ is given by substituting the values of $\vec{a}\ and\ \vec{n}$ , we get
p = $\left|\frac{2\vec{a}.(2\hat{i}+4\hat{j}+6\hat{k})+7}{|2\hat{i}+4\hat{j}+6\hat{k}|}\right|$
= $\left|\frac{2\vec{a}.(\hat{i}+2\hat{j}+3\hat{k})+7}{\sqrt{2^2+4^2+6^2}}\right|$
= $\left|\frac{2(-7)+7}{\sqrt{56}}\right|$
p = $\frac{7}{\sqrt{56}}$
Therefore, the distance between the planes
$\vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0$ and $\vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0$ is $\frac{7}{\sqrt{56}}$