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Question

Find the equation of the plane passing through the point (1,-1,2) having 2,3,2 as direction ratios of normal to the plane.

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Solution

The plane passes through the point having position vetor a=ˆiˆj+2ˆk and is normal to the vector n=2ˆi+3ˆj+2ˆk. So, the vector equation of the plane is
(ra).n=0
r.n=a.n
r.(2ˆi+3ˆj+2ˆk)=(ˆiˆj+2ˆk).(2ˆi+3ˆj+2ˆk)
r.(2ˆi+3ˆj+2ˆk)=3
The cartesian equation of the plane is 2x+3y+2z=3.

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