A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
E) \[\frac{2\pi }{3}\]
Correct Answer: D
Solution :
Given lines and planes are \[\frac{3x-1}{3}=\frac{y+3}{-1}=\frac{5-2z}{4}\] Or \[\frac{x-\frac{1}{3}}{1}=\frac{y+3}{-1}=\frac{\left( z-\frac{5}{2} \right)}{-2}\] and \[3x-3y-6z=0\] \[\Rightarrow \] \[x-y-2z=0\] Here, \[{{a}_{1}}=1,{{b}_{1}}=-1,{{c}_{1}}=-2\] and \[{{a}_{2}}=1,{{b}_{2}}=-1,{{c}_{2}}=-2\] \[\therefore \] \[\sin \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] \[=\frac{1\times 1+(-1)\times (-1)+(-2)\times (-2)}{\sqrt{1+1+4}\sqrt{1+1+4}}\] \[=\frac{6}{\sqrt{6}\sqrt{6}}=1\] \[\Rightarrow \] \[\theta =\frac{\pi }{2}\]
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