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) \[\frac{3\hat{i}-\hat{j}}{\sqrt{10}}\]
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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