Vector Algebra Question 84
Question: A vector of magnitude 3, bisecting the angle between the vectors $ \overset{\to }{\mathop{a}},=2\hat{i}+\hat{j}-\hat{k} $ and $ \overset{\to }{\mathop{b}},=\hat{i}-2\hat{j}+\hat{k} $ and making an obtuse angle with $ \overset{\to }{\mathop{b}}, $ is
Options:
A) $ \frac{3\hat{i}-\hat{j}}{\sqrt{6}} $
B) $ \frac{\hat{i}+3\hat{j}-2\hat{k}}{\sqrt{14}} $
C) $ \frac{3(\hat{i}+3\hat{j}-2\hat{k})}{\sqrt{14}} $
D) $ \frac{3\hat{i}-\hat{j}}{\sqrt{10}} $
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Answer:
Correct Answer: C
Solution:
- [c] A vector bisecting the angle between $ \overset{\to }{\mathop{a}}, $ and $ \overset{\to }{\mathop{b}}, $ is $ \frac{\overset{\to }{\mathop{a}},}{|\overset{\to }{\mathop{a}},|}\pm \frac{\overset{\to }{\mathop{b}},}{|\overset{\to }{\mathop{b}},|}; $ in this case $ \frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\pm \frac{\hat{i}-2\hat{j}+\hat{k}}{\sqrt{6}} $ i.e., $ \frac{3\hat{i}-\hat{j}}{\sqrt{6}}or\frac{\hat{i}+3\hat{j}-2\hat{k}}{\sqrt{6}} $ A vector of magnitude 3 along these vectors is $ \frac{3(3\hat{i}-\hat{j})}{\sqrt{10}} $ or $ \frac{3(\hat{i}+3\hat{j}-2\hat{k})}{\sqrt{14}} $ Now, $ \frac{3}{\sqrt{14}}(\hat{i}+3\hat{j}-2\hat{k}).(\hat{i}-2\hat{j}+\hat{k}) $ is negative and hence $ \frac{3}{\sqrt{14}}(\hat{i}+3\hat{j}-2\hat{k}) $ makes an obtuse angle with $ \overset{\to }{\mathop{b}}, $ .
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