A) 1
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{\sqrt{3}}\]
D) \[\frac{1}{2}\]
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 \[+vex-axis\] \[\therefore l=\cos \alpha \Rightarrow \cos \alpha =\frac{1}{\sqrt{3}}\]
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