Three Dimensional Geometry Question 301
Question: If $ \overset{\to }{\mathop{r}},=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}+\hat{j}+\hat{k}) $ and $ \overset{\to }{\mathop{r}},=(\hat{i}+2\hat{j}+3\hat{k})+\mu (\hat{i}+\hat{j}-\hat{k}) $ are two lines, then the equation of acute angle bisector of two lines is
Options:
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
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Answer:
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}}( \frac{(1)\times (1)+(-1)(1)+(1)(-1)}{\sqrt{3}\sqrt{3}} ) $ $ ={{\cos }^{-1}}( -\frac{1}{\sqrt{3}} ) $ Which is an obtuse angle.
$ \therefore $ Vector along acute angle bisector $ =\lambda [ \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}-\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}} ]=\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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