Question

The equation of the plane through the point (2,5,-3) perpendicular to the planes

Answer 1

Step-by-step explanation:

Toolbox:

A(x−x1)+B(y−y1)+C(z−z1)=0

Step 1:

Let the equation of the plane containing the given point (1,−1,2) is

A(x−x1)+B(y−y1)+C(z−z1)=0

(i.e) A(x−1)+B(y+1)+c(z−2)=0------(1)

Applying the condition of perpendicularly to the plane in equ(1) with the plane

2x+3y−2z=5 and x+2y−3z=8

We have 2A+3B−2C=0 and A+2B−3C=0

Step 2:

Let us solve three two equations,we get

A=−5C

B=4C

The required equation is

−5C(x−1)+4C(y+1)+c(z−2)=0

5x−4y−z−7=0

5x−4y−z=7