A) 0
B) \[{{\cos }^{-1}}\left[ \frac{18}{(15\sqrt{14})} \right]\]
C) \[{{\cos }^{-1}}\left[ \frac{-3}{\sqrt{20}} \right]\]
D) \[{{\sin }^{-1}}\left( \frac{2}{\sqrt{102}} \right)\]
E) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
Given that, \[\overrightarrow{r}=\hat{i}+2\hat{j}-\hat{k}+s(3\hat{i}-\hat{k})\] and \[\overrightarrow{r}=4\hat{i}-\hat{j}-3k+t(-4\hat{i}+\hat{j}+2\hat{i}+\hat{j})\] \[\Rightarrow \] \[\vec{r}=4\hat{i}-\hat{j}-3\hat{k}+t(-2\hat{i}+2\hat{j})\] \[\therefore \] \[\cos \theta =\frac{{{\overrightarrow{b}}_{1}}.{{\overrightarrow{b}}_{2}}}{|{{\overrightarrow{b}}_{1}}||{{\overrightarrow{b}}_{2}}|}\] \[=\frac{(3\hat{i}-\hat{k}).(-2\hat{i}+2\hat{j})}{\sqrt{{{(3)}^{2}}+{{(-1)}^{2}}}\sqrt{{{(-2)}^{2}}+{{(2)}^{2}}}}\] \[=\frac{-6}{\sqrt{9+1}\sqrt{4+4}}=\frac{-6}{\sqrt{80}}\] \[=-\frac{6}{2\sqrt{20}}=\frac{-3}{\sqrt{20}}\] \[\theta ={{\cos }^{-1}}\left( \frac{-3}{\sqrt{20}} \right)\]
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