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The equation of the plane perpendicular to the $z$-axis and passing through $ (2,-3,5) $ is
The equation of the plane perpendicular to the $z$-axis and passing through $ (2,-3,5) $ is
Updated On: Jun 23, 2024
- $ x-2=0 $
- $ y+3=0 $
- $ z-5=0 $
- $ 2x-3y+5z+4=0 $
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The Correct Option is C
Solution and Explanation
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 $
$ 0(x-2)+0(y+3)+1(z-5)=0 $ $ \Rightarrow $ $ z-5=0 $
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Concepts Used:
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Direction Cosines and Direction Ratios of Line:
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