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Co - ordinate Geometry 3D Geometry

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The equation of plane passing through the point (1, 1, 1), (1, –1, 1) and (–7, –3, –5) is perpendicular to

A. yz plane

B. xz plane

C. Either xz plane or yz plane

D. xy plane

E. None of the above

View Answer

Correct choice: D

Explanation:

Equation of a plane passing through the point (–7, –3, –5) is
a(x + 7) + b(y + 3) + c(x + 5) = 0
Since it passes through the points (1, 1, 1) and (1, –1, 1)
8a + 4b + 6c = 0 and 8a – 4b + 6c = 0
Solving we get a/48 = b/0 = c/-64
∴d.r. of normal to the plane is 3, 0, –4 hence is perpendicular to xz plane

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Eamcet Mathematics Co - ordinate Geometry 3D Geometry

Eamcet Mathematics
Co - ordinate Geometry 3D Geometry