Suppose \(\rm \mathop v\limits^ \to = 2\hat i + \hat j - \hat k\)  and \(\rm \mathop w\limits^ \to = \hat i + 3\hat k\). If \(\rm \mathop u\limits^ \to \) is unit vector, then the maximum value of scalar triple product \(\rm \left[ {\mathop u\limits^ \to \mathop v\limits^ \to \mathop w\limits^ \to } \right]\) is -

Concept:

Scalar Triple Product: 

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\), 

\(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) and,

\(\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\)

Then their scalar triple product is defined as,

\(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

Calculation:

Given:

We have,

\(\rm \mathop v\limits^ \to = 2\hat i + \hat j - \hat k\)

\(\rm \mathop w\limits^ \to = \hat i + 3\hat k\)

\(\rm \vec v \times \vec w = \begin{bmatrix} \hat i & \hat j & \hat k \\\ 2 & 1 & -1 \\\ 1 & 0 & 3 \end{bmatrix}\)

\(\Rightarrow \vec v \times \vec w = \hat i (3) - \hat j (7) + \hat k (-1)\)

\(\rm \Rightarrow \vec v \times \vec w = 3\hat i - 7\hat j - \hat k \)

Now, \(\rm \vec u . (\vec v \times \vec w) = | \vec u| |(3\hat i - 7\hat j - \hat k )| \cos θ \)

If \(\rm \mathop u\limits^ \to \) is a unit vector so, 

\(\rm \vec u . (\vec v \times \vec w) = 1 |(3\hat i - 7\hat j - \hat k )| \cos θ \)

When, θ = 0°

Maximum value = \(\sqrt{3^2 +(-7)^2 + (-1)^2 } = \sqrt{59}\)