Question

Find the equation of plane which passes through the points (1.-2. 4) and (3.-4,5) and perpendicular to the plane x + y - 22 - 6​

Answer 1

Answer:

Equation of the plane passing through (1,−2,4) is A(x−1)+B(y+2)+C(z−4)=0 ---(1)

Since this plane passes through (3,−4,5)

So,

A(3−1)+B(−4+2)+C(5−4)=0

⇒2A−2B+C=0 ---(2)

also this plane is perpendicular to yz-plane i.e plane x=0

A×1+B×0+C×0=0

⇒A=0

Putting A=0 in (2) we get

0−2B+C=0

⇒2B=C

⇒B=

2

C

Putting values of A and B in (1) we get

0(x−1)+

2

C

(y+2)+C(Z−4)=0

⇒y+2+2(z−4)=0

⇒y+2+2z−8=0

⇒y+2z−6=0

The required equation of plane is y+2z−6=0