Question

Find the equation of the planes parallel to the plane x-2y+2z-3=0 whose perpendicular distance from (1,2,3) is 1.

Answer 1

any plane parallel to \(x-2y+2z-3 = 0\) willl be of form

\(x-2y+2x+k=0\)

Answer 2

since you're given the \(\perp\) distance from point (1, 2, 3), there can be two planes i think
you can find them

Answer 3

how can i find?

Answer 4

\(\perp \) distance from a point \((x_1, y_1, z_1)\) to a plane \(ax+by+cz+d = 0\) is

\(\large \frac{ax_1 + by_1+cz_1 + d}{\sqrt{a^2+b^2+c^2}}\)

Answer 5

set it to 1 and find k

Answer 6

ok and then i have to put the value of k in dat eqn right?

Answer 7

ok m wrking i tout

Answer 8

yes

Answer 9

thank u! :)

Answer 10

np :) think a bit more.... im not sure whether there will be only 1 plane or 2 planes

Answer 11

hey m getting k=root14-3