Vectors Ques 85
- If $\overrightarrow{\mathbf{V}}=2 \overrightarrow{\mathbf{i}}+\overrightarrow{\mathbf{j}}-\overrightarrow{\mathbf{k}}$ and $\overrightarrow{\mathbf{W}}=\overrightarrow{\mathbf{i}}+3 \overrightarrow{\mathbf{k}}$. If $\overrightarrow{\mathbf{U}}$ is a unit vector, then the maximum value of the scalar triple product $[\overrightarrow{\mathbf{U}} \overrightarrow{\mathbf{V}} \overrightarrow{\mathbf{W}}]$ is
(2002, 1M)
(a) -1
(b) $\sqrt{10}+\sqrt{6}$
(c) $\sqrt{59}$
(d) $\sqrt{60}$
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Answer:
Correct Answer: 85.(c)
Solution:
Formula:
- Given, $\overrightarrow{\mathbf{V}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{W}}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$
$$ \begin{aligned} {[\overrightarrow{\mathbf{U}} \overrightarrow{\mathbf{V}} \overrightarrow{\mathbf{W}}] } & =\overrightarrow{\mathbf{U}} \cdot[(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \times(\hat{\mathbf{i}}+3 \hat{\mathbf{k}})] \\ & =\overrightarrow{\mathbf{U}} \cdot(3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}})=|\overrightarrow{\mathbf{U}}||3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}| \cos \theta \end{aligned} $$
which is maximum, if angle between $\overrightarrow{\mathbf{U}}$ and $3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ is 0 and maximum value
$$ =|3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}|=\sqrt{59} $$
BETA
