13. 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.
two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
the normal vectors of these plane are
Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :
Now, as we know
the equation of a plane in vector form is :
Now Since this plane passes through the point (-1,3,2)
Hence the equation of the plane is
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