A) \[\overset{\to }{\mathop{r}}\,=(\hat{i}+2\hat{j}+3\hat{k})+t(\hat{j}-\hat{k})\]
B) \[\overset{\to }{\mathop{r}}\,=(\hat{i}+2\hat{j}+3\hat{k})+t(2\hat{i})\]
C) \[\overset{\to }{\mathop{r}}\,=(\hat{i}+2\hat{j}+3\hat{k})+t(\hat{j}+\hat{k})\]
D) None of these
Correct Answer: A
Solution :
[a] Lines are \[\vec{r}=(\hat{i}++2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})\] And \[\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\mu (\hat{i}+\hat{j}-\hat{k})\] Along vectors \[(\hat{i}-\hat{j}+\hat{k})\] and \[(\hat{i}+\hat{j}-\hat{k})\] Respectively. Angle between two lines \[={{\cos }^{-1}}\left( \frac{(1)\times (1)+(-1)(1)+(1)(-1)}{\sqrt{3}\sqrt{3}} \right)\] \[={{\cos }^{-1}}\left( -\frac{1}{\sqrt{3}} \right)\] Which is an obtuse angle. \[\therefore \] Vector along acute angle bisector \[=\lambda \left[ \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}-\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}} \right]=\frac{2\lambda }{\sqrt{3}}(-\hat{j}+\hat{k})\] \[\therefore \] Equation of acute angle bisector \[=(\hat{i}+2\hat{j}+3\hat{k})+t(\hat{j}-\hat{k})\]
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