A) \[{{\cos }^{-1}}\left( \frac{1}{abc} \right)\]
B) \[{{\sin }^{-1}}\left( \frac{1}{abc} \right)\]
C) \[{{0}^{o}}\]
D) \[{{90}^{o}}\]
Correct Answer: D
Solution :
Key Idea: If \[{{a}_{1}},{{b}_{2}},{{c}_{1}}\]and \[{{a}_{2}},{{b}_{2}},{{c}_{2}}\]are the direction ratios of the lines, then \[\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] Since, \[a,\text{ }b,\text{ }c\]and \[b-c,c-a,a-b\]are direction ratios of the lines, then the angle between lines \[\theta ={{\cos }^{-1}}\frac{a.(b-c)+b.(c-a)+c.(a-b)}{\left[ \begin{align} & \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \\ & \times \sqrt{{{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(a-b)}^{2}}} \\ \end{align} \right]}\] \[={{\cos }^{-1}}(0)=\frac{\pi }{2}\]
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