Question

find the vector equation passing through the point(2,2,3) and 3,4,3 as direction ratios normal to the plane
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Answer 1

The plane passes through the point having position vetor a=i−j+2k and is normal to the vector n=2i+3j+2k. So, the vector equation of the plane is

(r−a).n=0

r.n=a.n

r.(2i+3j+2k)=(i−j+2k).(2i+3j+2k)

r.(2i+3j+2k)=3

The cartesian equation of the plane is 2x+3y+2z=3.