Question

Let P be the plane, which contains the line of intersection of the planes $x+y+z-6=0$ and $2x+3y+z+5=0$, and it is perpendicular to the xy-plane. Find the distance of the point $\left( 0,0,256 \right)$ from the P plane?
(a) $63\sqrt{5}$,
(b) $205\sqrt{5}$,
(c) $\dfrac{17}{\sqrt{5}}$,
(d) $\dfrac{11}{\sqrt{5}}$.

Answer 1

We start solving the problem by assuming a general equation for the required plane P. We use the fact that the equation of xy-plane is $z=0$ to use the condition that P is perpendicular to xy-plane. We then find the general equation of the plane that contains the line of intersection of the planes $x+y+z-6=0$ and $2x+3y+z+5=0$ to proceed further into the problem. We then compare both equations and get the equation of the plane P. Using this equation, we then find the required distance.

Complete step-by-step answer :
According to the problem we have a plane P, which contains the line of intersection of the planes $x+y+z-6=0$ and $2x+3y+z+5=0$. The plane P is perpendicular to xy-plane. We need to find the distance of point $\left( 0,0,256 \right)$ from the plane P.
We know that the equation of the xy-plane is $z=0$. Let us assume the equation of the plane P be $ax+by+cz+d=0$.
We know that if two planes $mx+ny+oz+p=0$ and $qx+ry+sz+t=0$ are perpendicular, then $mq+nr+os=0$. We use this result for the planes $z=0$ and $ax+by+cz+d=0$.
$\Rightarrow \left( 0\times a \right)+\left( 0\times b \right)+\left( 1\times c \right)=0$.
$\Rightarrow c=0$.
So, we have got the equation of the plane P as $ax+by+d=0$ ---(1).
We know that the equation of the plane having intersection of two other planes ${{P}_{1}}$ and ${{P}_{2}}$ is ${{P}_{1}}+\alpha {{P}_{2}}=0$.
We have equation of the plane P as $\left( x+y+z-6 \right)+\alpha \left( 2x+3y+z+5 \right)=0$.
$\Rightarrow x+y+z-6+2\alpha x+3\alpha y+\alpha z+5\alpha =0$.
$\Rightarrow \left( 1+2\alpha \right)x+\left( 1+3\alpha \right)y+\left( 1+\alpha \right)z+\left( 5\alpha -6 \right)=0$ ---(2).
Comparing equations (1) and (2), we get
$\Rightarrow 1+\alpha =0$.
$\Rightarrow \alpha =-1$ ---(3).
We substitute equation (3) in equation (2).
$\Rightarrow \left( 1+2\left( -1 \right) \right)x+\left( 1+3\left( -1 \right) \right)y+\left( 1+\left( -1 \right) \right)z+\left( 5\left( -1 \right)-6 \right)=0$.
$\Rightarrow \left( 1-2 \right)x+\left( 1-3 \right)y+\left( 1-1 \right)z+\left( -5-6 \right)=0$.
$\Rightarrow -x-2y-11=0$.
$\Rightarrow x+2y+11=0$ ---(4).
We need to find the distance of point $\left( 0,0,256 \right)$ from the plane $x+2y+11=0$.
We know that the distance of the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ from the plane $px+qy+rz+s=0$ is $\dfrac{\left| p{{x}_{1}}+q{{y}_{1}}+r{{z}_{1}}+s \right|}{\sqrt{{{p}^{2}}+{{q}^{2}}+{{r}^{2}}}}$.
We have distance of point $\left( 0,0,256 \right)$ from the plane $x+2y+11=0$ as $\dfrac{\left| 0+2\left( 0 \right)+11 \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{0}^{2}}}}$.
$\Rightarrow \dfrac{\left| 0+11 \right|}{\sqrt{1+4+0}}$.
$\Rightarrow \dfrac{\left| 11 \right|}{\sqrt{5}}$.
$\Rightarrow \dfrac{11}{\sqrt{5}}$.
We have found the distance of point $\left( 0,0,256 \right)$ from the plane P as $\dfrac{11}{\sqrt{5}}$.
∴ The distance of point $\left( 0,0,256 \right)$ from the plane P is $\dfrac{11}{\sqrt{5}}$.
The correct option for the given problem is (d).

Note : We can solve the problem by finding the direction ratios of the line of intersection of planes which can be used to find the equation of line. We should not confuse what to multiply 256 with as there is 0z present here. We can add 0z to the equation of the plane to avoid confusion. Similarly, we can expect problems to find the direction ratios and cosines of the plane and line of intersection.