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  4. The equation of the plane perpendicular to the z-axis and passing through (2,-3,5) is

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Q. The equation of the plane perpendicular to the $z$-axis and passing through $ (2,-3,5) $ is

J & K CETJ & K CET 2011Three Dimensional Geometry

Solution:

Since, the plane perpendicular to z-axis, so the normal of the plane is parallel to z-axis, the direction cosine's of z-axis is $ (0,0,1) $ . Then, the equation of plane which passes through the point $ (3,-3,5) $ and perpendicular to z-axis is
$ 0(x-2)+0(y+3)+1(z-5)=0 $ $ \Rightarrow $ $ z-5=0 $