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Question: Answered & Verified by Expert
Let $\mathbf{V}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{W}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$. If $\mathbf{U}$ is a unit vector, then the maximum value of $[\mathbf{U} \mathbf{V} \mathbf{W}]$ is
MathematicsVector AlgebraTS EAMCETTS EAMCET 2020 (14 Sep Shift 1)
Options:
  • A -1
  • B $\sqrt{10}+\sqrt{16}$
  • C $\sqrt{59}$
  • D $\sqrt{60}$
Solution:
1830 Upvotes Verified Answer
The correct answer is: $\sqrt{59}$
$\begin{aligned} \text { Given, } \mathbf{V} & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \mathbf{W} & =\hat{\mathbf{i}}+3 \hat{\mathbf{k}} \\ \mathbf{V} \times \mathbf{W} & =\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -1 \\ 1 & 0 & 3\end{array}\right|=3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}\end{aligned}$
Here, $\mathbf{U}$ is unit vector
$\begin{aligned}
\therefore \text { Maximum value of } \mathbf{U} \cdot(\mathbf{V} & \times \mathbf{W})=|\mathbf{U}| \cdot|\mathbf{V} \times \mathbf{W}| \\
& =1 \times \sqrt{9+49+1}=\sqrt{59}
\end{aligned}$

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