Three Dimensional Geometry Question 135
Question: Let L be the line of intersection of the planes $ 2x+3y+z=1 $ and $ x+3y+2z=2 $ . If L makes an angle $ \alpha $ with the positive x-axis, then $ cos\alpha $ equals
Options:
A) 1
B) $ \frac{1}{\sqrt{2}} $
C) $ \frac{1}{\sqrt{3}} $
D) $ \frac{1}{2} $
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Answer:
Correct Answer: C
Solution:
[c] Let the direction cosines of line L be l, m, n, then $ 2l+3m+n=0 $ (i) And $ l+3m+2n=0 $ (ii) On solving equations (i) and (ii), we get $ \frac{l}{6-3}=\frac{m}{1-4}=\frac{n}{6-3}\Rightarrow \frac{l}{3}=\frac{m}{-3}=\frac{n}{3} $ Now $ \frac{l}{3}=\frac{m}{-3}=\frac{n}{3}=\frac{\sqrt{l^{2}+m^{2}+n^{2}}}{\sqrt{3^{2}+{{(-3)}^{2}}+3^{2}}} $
$ \therefore l^{2}+m^{2}+n^{2}=1\therefore \frac{l}{3}=\frac{m}{-3}=\frac{n}{3}=\frac{1}{\sqrt{27}} $
$ \Rightarrow l=\frac{3}{\sqrt{27}}=\frac{1}{\sqrt{3}},m=-\frac{1}{\sqrt{3}},n=\frac{1}{\sqrt{3}} $ Line L, makes an angle $ \alpha $ with $ +vex-axis $
$ \therefore l=\cos \alpha \Rightarrow \cos \alpha =\frac{1}{\sqrt{3}} $
BETA
