Question

Find the equation of plane passing through (2,0,1)and(3,-3,4)and perpendicular to x-2y+z=6

Answer 1

Answer:

Step-by-step explanation:

equation of plane from two points is

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

also second point lies on it so

a(3-2)+b(-3-0)+c(4-1)=0

you will get a-3b+3c=0....(1)

and this plane is perpendicular to given palne so product of direction ratios is zero i.e

a(1)+b(-2)+c(1)=0

a-2b+c=0 ........(2)

solve 1 and 2

from cross multiplication and find a,b,c put back in *

that is your answer

Answer 2

Step-by-step explanation:

this is the equation of required plane