A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) None of these
Correct Answer: D
Solution :
Given line can be rewritten as\[\frac{x-\frac{1}{3}}{1}=\frac{y+3}{-1}=\frac{\left( z-\frac{5}{2} \right)}{-2}\]and equation of plane is rewritten as \[x-y-2z=\frac{10}{3}\] Here, direction ratios of line and plane are\[{{a}_{1}}=1,{{b}_{1}}=-1\] and \[{{c}_{1}}=-2\]and \[{{a}_{2}}=1,{{b}_{2}}=-1\]and \[{{c}_{2}}=-2\]x Now, \[\sin \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}}\] \[=\frac{1\times 1+(-1)\times (-1)+(-2)\times (-2)}{\sqrt{1+1+4}\sqrt{1+1+4}}\] \[=\frac{1+1+4}{\sqrt{6}\sqrt{6}}=\frac{6}{6}=1\] \[\Rightarrow \] \[\theta =\frac{\pi }{2}\]
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