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Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Asked by Reshma Last Modified
Answer The equation of the plane passing through the point (−1, 3, 2) is
a (x + 1) + b (y − 3) + c (z − 2) = 0 … (1)
where, a, b, c are the direction ratios of normal to the plane.
It is known that two planes, 
and 
, are perpendicular, if 
Plane (1) is perpendicular to the plane, x + 2y + 3z = 5
Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0
From equations (2) and (3), we obtain
Substituting the values of a, b, and c in equation (1), we obtain
This is the required equation of the plane.
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