Answer:
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 =[tex](2-3, -1-4, 5-(-1))\\=(-1, -5, 6)[/tex]
this is direction ratios of normal
The plane passes through the point (3,-3,1)
Hence equation of the plane in normal form is
[tex]-1(x-3) -5(y+3)+6(z-1) =0[/tex]
[tex]-1x+3 -5y-15+6z-6 =0[/tex]
x+5y-6z+18 =0 is the equation of the plane.
