Three Dimensional Geometry Question 117
Question: The direction cosines of the normal to the plane $ x+2y-3z+4=0 $ are
[MP PET 1996; Pb. CET 2000]
Options:
A) $ -\frac{1}{\sqrt{14}},-\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}} $
B) $ \frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}} $
C) $ -\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}} $
D) $ \frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}} $
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Answer:
Correct Answer: A
Solution:
The direction cosines of the normal to the plane are $ \frac{1}{\sqrt{1^{2}+2^{2}+3^{2}}},\frac{2}{\sqrt{1^{2}+2^{2}+3^{2}}},\frac{-,3}{\sqrt{1^{2}+2^{2}+3^{2}}} $ i.e., $ \frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{-,3}{\sqrt{14}} $ . But $ x+2y-3z+4=0 $ can be written as $ -x-2y+3z-4=0 $ . Thus the direction cosines are $ \frac{-1}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{3}{\sqrt{14}} $ .
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