Question

Calculus 3 Planes Review

Answer 1

Given equation of planes are x+y+z=6,2x+3y+4z+5=0

Equation of plane passing through intersection of given plane is,

L

1

+λL

2

=0⇒x+y+z−6+λ(2x+3y+4z+5)=0

Given, it also passes through (1,1,1)

⇒1+1+1+−6+λ(2+3+4+5)=0⇒λ=

14

3

Hence, required plane is, 14(x+y+z−6)+3(2x+3y+4z+5)=0

⇒20x+23y+26z−69=0