Question

If the equation of a plane $P$, passing through the intersection of the planes, $x+4y-z+7=0$ and $3x+y+5z=8$ is $ax+by+6z=15$ for some $a,b\in R$, then the distance of the point$(3,2,-1)$ from the plane $P$ is ________ units.

Answer 1

Step 1: Find the equation of the plane

Let $P_1$ and $P_2$ be two planes.

Say $P_1:x+4y-z+7=0$ and $P_2:3x+y+5z=8$

Equation of plane passing through line of intersection of given planes is given by

$$\begin{gathered} P_1+\lambda P_2=0 \\ \Rightarrow \left(x+4y-z+7\right)+\lambda \left(3x+y+5z-8\right)=0 \\ \Rightarrow x\left(1+3\lambda \right)+y\left(4+\lambda \right)+z\left(\lambda 5-1\right)+\left(7-8\lambda \right)=0....\left(i\right) \end{gathered}$$

Step 2: Comparing the equation of the plane with the given equation

The given plane $P$ is $ax+by+6z=15.....\left(ii\right)$

Comparing equation $\left(i\right)$ and $\left(ii\right)$ we get,

$\frac{1+3\lambda }{a}=\frac{4+\lambda }{b}=\frac{-1+5\lambda }{6}=\frac{7-8\lambda }{-15}$

From $\frac{-1+5\lambda }{6}=\frac{7-8\lambda }{-15}$ we have

$$\begin{gathered} \Rightarrow 15-75\lambda =42-48\lambda \\ \Rightarrow 27\lambda =-27 \\ \Rightarrow \lambda =-1 \end{gathered}$$

From $\frac{1+3\lambda }{a}=\frac{-1+5\lambda }{6}$ we get

$$\begin{gathered} \Rightarrow \frac{1+3\lambda }{a}=\frac{-1+5\lambda }{6} \\ \Rightarrow \frac{1+3\left(-1\right)}{a}=\frac{-1+5\left(-1\right)}{6} \\ \Rightarrow a=2 \end{gathered}$$

From $\frac{4+\lambda }{b}=\frac{-1+5\lambda }{6}$ we get

$$\begin{gathered} \Rightarrow \frac{4+\lambda }{b}=\frac{-1+5\lambda }{6} \\ \Rightarrow \frac{4-1}{b}=\frac{-1+5\left(-1\right)}{6} \\ \Rightarrow \frac{3}{b}=\frac{-6}{6} \\ \Rightarrow b=-3 \end{gathered}$$

Therefore, the equation of the plane is $2x-3y+6z-15=0$

Step 3: Find the distance of the plane from the point

We know that the distance of the point $\left(x_1,y_1,z_1\right)$ from the plane $ax+by+cz+d=0$ is $\frac{\left|a{x}_1+b{y}_1+c{z}_1+d\right|}{\sqrt{a^2+b^2+c^2}}$

Therefore the distance of the point $(3,2,-1)$ from the plane $2x-3y+6z-15=0$ is

Distance$=\frac{\left|2\left(3\right)-3\left(2\right)+6\left(-1\right)-15\right|}{\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2+{\left(6\right)}^2}}$

$$\begin{gathered} =\frac{\left|-21\right|}{\sqrt{49}} \\ =3\mathrm{unit}s \end{gathered}$$

Hence, the distance of the point$(3,2,-1)$ from the plane $P$ is $3\mathrm{unit}s$