Three Dimensional Geometry Question 268
Question: What are the direction cosines of a line which is equally inclined to the positive directions of the axes?
Options:
A) $ \langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \rangle $
B) $ \langle -\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \rangle $
C) $ \langle -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \rangle $
D) $ \langle \frac{1}{3},\frac{1}{3},\frac{1}{3} \rangle $
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Answer:
Correct Answer: A
Solution:
[a] Let $ \ell $ , m, n are the dc?s of a line that is inclined equally at $ \alpha $ to the +ve direction of axes. Now, $ \ell =\cos \alpha ,m=\cos \alpha .n=\cos \alpha . $ Also, $ {{\ell }^{2}}+m^{2}+n^{2}=1 $ $ 3{{\cos }^{2}}\alpha =1. $ $ \cos \alpha =\frac{1}{\sqrt{3}} $
$ \therefore $ dc?s of the line are: $ \langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} \rangle $
BETA
