A) \[\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{2}\]
B) \[\frac{1}{\sqrt{2}},-\frac{1}{2},\frac{1}{2}\]
C) \[-\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\]
D) \[\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\]
Correct Answer: C
Solution :
Let \[m=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\] and \[n=\cos \frac{\pi }{3}=\frac{1}{2}\] \[\because \] \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\] \[\Rightarrow \] \[l=\sqrt{1-({{m}^{2}}+{{n}^{2}})}=\sqrt{1-\left( \frac{1}{2}+\frac{1}{4} \right)}\] \[=\sqrt{1-\frac{3}{4}}\] \[\Rightarrow \] \[l=\pm \frac{1}{2}\] Since, line makes an obtuse angle, so we angle \[l=-\frac{1}{2}\] \[\therefore \] Direction cosines are \[-\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2},\]
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