Question

Find the equation of the plane through the line of intersection of the planes x-z = 3 and y+3z = 4 and perpendicular to the plane x+y+z = 1.

Answer 1

ANSWER:-

Given that

the planes x-z = 3 and y+3z = 4

and perpendicular to the plane x+y+z = 1.

The lanes - Z=3 and y + 32=4 . perpendicular to the plane at yt2=1 Equation of a plane which passes through intersection Pino and P2 = 0 is of The form P, 4x12=0 Piti1220 (2-2-3) + 1 (y +32-4)=0 2- Ayo (11) ? – (3 +44)=0 (4-1) Z-(3+4x)=6 - 0 . Two planes are perpendicular only if thein. normal vectors are perpendicwar > ALSO dot produit of normal vector is zero. so we can write. (1,2, 3-1) (lila) ... 1-1-1-1=0 - 2x=0

substitute ge the value & in eq 0 to the equation of Plane. x-olyst. (0-1) Z- (3 + 4 (0)=0 - 1-2-3-0.