Question

If $\overrightarrow{V}=2\vec{i}+\vec{j}-\overrightarrow{k}$ and $\overrightarrow{W}=\overrightarrow{i}+3\overrightarrow{k}$. If $\overrightarrow{U}$ is a unit vector, then the maximum value of the scalar triple product $[\overrightarrow{U}\overrightarrow{V}\overrightarrow{W}]$ is

• A. $-1$
• B. $\sqrt{10}+\sqrt{6}$
• C. $\sqrt{59}$
• D. $\sqrt{60}$

Answer 1

Answer: (C)

Given, $\overrightarrow{V}=2\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{w}=\hat{i}+3\hat{k}$
$[\overrightarrow{U}\overrightarrow{V}\overrightarrow{W}]=\overrightarrow{U}\cdot [(2\hat{i}+\hat{j}-\hat{k})\times (\hat{i}+3\hat{k})]$
$=\overrightarrow{U}\cdot (3\hat{i}-7\hat{j}-\hat{k})=|\overrightarrow{U}||3\hat{i}-7\hat{j}-\hat{k}|\cos \theta$
which is maximum, if angle between $\vec{U}$ and $3\hat{i}-7\hat{j}-\hat{k}$ is $0$ and maximum value
$=|3\hat{i}-7\hat{j}-\hat{k}|=\sqrt{59}$