Question

Find the distance of the point (1, $-$2, 4) from plane passing throuhg the point (1, 2, 2) and perpendicular of the planes $x-y+$2$z=$3 and 2$x-$2$y+z+$12$=$0.

Answer 1

Let the equation of plane passing through the point (1, 2, 2) be

$a\left(x-1\right)+b\left(y-2\right)+c\left(z-2\right)=0$ .....(1)

Here, a, b, c are the direction ratios of the normal to the plane.

The equations of the given planes are x − y + 2z = 3 and 2x − 2y + z + 12 = 0.

Plane (1) is perpendicular to the given planes.

a − b + 2c = 0 .....(2)

2a − 2b + c = 0 .....(3)

Eliminating a, b and c from (1), (2) and (3), we get

$$\begin{gathered} \left|\begin{array}{ccc}x-1& y-2& z-2\\ 1& -1& 2\\ 2& -2& 1\end{array}\right|=0 \\ \Rightarrow \left(x-1\right)\left(-1+4\right)-\left(y-2\right)\left(1-4\right)+\left(z-2\right)\left(-2+2\right)=0 \\ \Rightarrow 3x+3y-9=0 \\ \Rightarrow x+y-3=0 \end{gathered}$$
∴ Distance of the point (1, −2, 4) from the plane x + y − 3 = 0

$$\begin{gathered} =\left|\frac{1-2-3}{\sqrt{1^2+1^2+0^2}}\right| \\ =\left|\frac{-4}{\sqrt{2}}\right| \\ =2\sqrt{2}\mathrm{units} \end{gathered}$$