Question

Find the vector equation of a plane passing through the point (1, 4, 6) and normal to the vector i-2j+k.

Answer 1

equation of plane when a ,b and c are direction cosine and (r,s ,t) is the point in the plane .

is ---------

a (x - r) + b (y - s) + (z - t) =0

now ,
here i -2j + k is normal to the plane so, direction cosine are 1 , -2 and 1

1 ( x -1) -2 (y - 4) + 1(z - 6) =0

x -2y + z + 1 =0