Question

The distance of the point $(1,-2,4)$ from the plane passing through the point $(1,2,2)$ and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $2\sqrt{2}$
• C. $2$
• D. $\sqrt{2}$

Answer 1

Answer: (B)

$2\sqrt{2}$

Let the equation of the plane be $ax+by+cz=d$

then point $(1,2,2)$ lies on the plane

So, $a+2b+2c=d$ .......eqn 1

Since the plane is perpendicular to the given two planes, so dot product of the direction ratios of the perpendicular of the plane with the direction ratio of the perpendicular of the given two plane must be zero.

So,
$a-b+2c=0$ and .......eqn 2

$2a-2b+c=0$ ........eqn 3

Doing eqn 1$-$eqn 2, we get

$3b=d$,

Similarly,

On solving above 3 eqn we get

$c=0$ and $a=b$ and $d=3a$

Put the value of $b,c$ and $d$ in terms of $a$ and get the equation of the plane as

$x+y=3$

Hence $d=\frac{|1-2-3|}{\sqrt{2}}=2\sqrt{2}$