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
$=|3 \hat{ i }-7 \hat{ j }-\hat{ k }|=\sqrt{59}$