The distance of the point (1,-2,4) from the plane passing through the point (1,2,2) and pe…

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

Accepted Answer

Equation of plane passing through point (1, 2, 2) is,

a(x - 1) + b(y - 2) + c(z - 2) = 0        . . .(1)

This plane is perpendicular to the plane

x - y + 2z = 3   and  2x - 2y + z + 12 = 0

When two planes,

a1x + b1y + c1z + d1 = 0

and a2x + b2y + c2z + d2 = 0

are perpendicular to each other then

a1a2 + b1b2 + c1c2 = 0

So, we can say,

a $$ \times $$ 1 + b $$ \times $$ (- 1) + c $$ \times $$ 2 = 0

$$ \Rightarrow $$   a - b + 2c = 0           . . . (2)

and, a $$ \times $$ 2 + b(-2) + c(1) = 0

$$ \Rightarrow $$   2a - 2b + c = 0           . . .(3)

Solving (2) and (3)

$${a \over { - 1 + 4}}$$ = $${b \over {4 - 1}}$$ = $${c \over { - 2 + 2}}$$ = $$\lambda $$

$$ \Rightarrow $$    a = 3$$\lambda $$, b = 3$$\lambda $$, c = 0

Putting the values of a, b, c in equation (1) we get,

3$$\lambda $$ (x - 1) + 3$$\lambda $$ (y - 2) = 0

$$ \Rightarrow $$   3(x - 1) + 3(y - 2) = 0

$$ \Rightarrow $$   3x + 3y - 9 = 0

$$ \Rightarrow $$   x + y - 3 = 0

$$ \therefore $$   Distance of point (1, -2,  4) from plane x + y - 3 = 0 is,

= $$\left| {{{1 - 2 - 3} \over {\sqrt {1 + 1} }}} \right|$$ = $${4 \over {\sqrt 2 }}$$ = 2$$\sqrt 2 $$

Solutions · 1