Three Dimensional Geometry Question 122
Question: The direction cosines of the normal to the plane $ 3x+4y+12z=52 $ will be
[MP PET 1997]
Options:
A) 3, 4, 12
B) ? 3, ? 4, ? 12
C) $ \frac{3}{13},\frac{4}{13},\frac{12}{13} $
D) $ \frac{3}{\sqrt{13}},\frac{4}{\sqrt{13}},\frac{12}{\sqrt{13}} $
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Answer:
Correct Answer: C
Solution:
Direction ratio of normal to the plane $ 3x+4y+12z=52 $ is 3, 4, 12.
$ \therefore $ D.c’s are $ ( \frac{3}{\sqrt{3^{2}+4^{2}+12^{2}}},,\frac{4}{\sqrt{3^{2}+4^{2}+12^{2}}},\frac{12}{\sqrt{3^{2}+4^{2}+12^{2}}} ) $ $ =( \frac{3}{13},\frac{4}{13},\frac{12}{13} ) $ .
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