Find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5).

x+5y-6z+18 =0

Step-by-step explanation:

We have to find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points A(3, 4, – 1) and B(2, – 1, 5).

The perpendicular line joining AB is normal to the plane.

Direction ratios of AB are =

this is direction ratios of normal

The plane passes through the point (3,-3,1)

Hence equation of the plane in normal form is

x+5y-6z+18 =0 is the equation of the plane.

where are the options?

step-by-step explanation:

step-by-step explanation: