Q73 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

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

  • A
    -1
  • B
    $\sqrt{10}+\sqrt{16}$
  • Correct answer
    C
    $\sqrt{59}$
  • D
    $\sqrt{60}$

Solution

$\begin{aligned} \text { Given, } \mathbf{V} & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \mathbf{W} & =\hat{\mathbf{i}}+3 \hat{\mathbf{k}} \\ \mathbf{V} \times \mathbf{W} & =\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -1 \\ 1 & 0 & 3\end{array}\right|=3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}\end{aligned}$ Here, $\mathbf{U}$ is unit vector $\begin{aligned} \therefore \text { Maximum value of } \mathbf{U} \cdot(\mathbf{V} & \times \mathbf{W})=|\mathbf{U}| \cdot|\mathbf{V} \times \mathbf{W}| \\ & =1 \times \sqrt{9+49+1}=\sqrt{59} \end{aligned}$

Hence, the correct answer is option: 'C'.

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