A) \[\sqrt{3}\]
B) \[5\]
C) \[\sqrt{14}\]
D) \[7\]
Correct Answer: B
Solution :
Let\[\overset{\to }{\mathop{u}}\,=\hat{j}+4\hat{k},\overset{\to }{\mathop{v}}\,=\hat{i}-3\hat{k}\]and\[\overset{\to }{\mathop{w}}\,=\cos \theta \hat{i}+\sin \theta \hat{j}\] Now,\[\overset{\to }{\mathop{u}}\,\times \overset{\to }{\mathop{v}}\,=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 0 & 1 & 4 \\ 1 & 0 & -3 \\ \end{matrix} \right|\] \[=\hat{i}\left( -3 \right)-\hat{j}\left( -4 \right)+\hat{k}\left( -1 \right)\] \[=-3\hat{i}+4\hat{j}-\hat{k}\]Now, \[\left( \overset{\to }{\mathop{u}}\,\times \overset{\to }{\mathop{v}}\, \right).\overset{\to }{\mathop{w}}\,=\left( -3\hat{i}+4\hat{j}-\hat{k} \right).\left( \cos \theta \hat{i}+\sin \theta \hat{j} \right)\] \[=-3\cos \theta +4\sin \theta \] Now, maximum possible value of \[|-3\cos \theta +4\sin \theta |=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 4 \right)}^{2}}}=\sqrt{25}=5\]
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