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 + 2z = 3 and 2x - 2y + 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

\[\begin{vmatrix}x - 1 & y - 2 & z - 2 \\ 1 & - 1 & 2 \\ 2 & - 2 & 1\end{vmatrix} = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( - 1 + 4 \right) - \left( y - 2 \right)\left( 1 - 4 \right) + \left( z - 2 \right)\left( - 2 + 2 \right) = 0\]
\[ \Rightarrow 3x + 3y - 9 = 0\]
\[ \Rightarrow x + y - 3 = 0\]

∴ Distance of the point (1, −2, 4) from the plane x + y − 3 = 0

\[= \left| \frac{1 - 2 - 3}{\sqrt{1^2 + 1^2 + 0^2}} \right|\]
\[ = \left| \frac{- 4}{\sqrt{2}} \right|\]
\[ = 2\sqrt{2} \text{ units } \]