Correct Answer - Option 1 : 1
Concept:
If \(\rm \vec{a}=a_{1}\hat i+a_{2}\hat j+a_{3}\hat k\), \(\rm \vec{b}=b_{1}\vec i+b_{2}\vec j+b_{3}\vec k\) and \(\rm \vec{c}=c_{1}\vec i+c_{2}\vec j+c_{3}\vec k\), then \(\rm \vec{a}.( \vec{b}\times \vec{c})= \begin{vmatrix} a_{1} &a_{2} &a_{3}\\ b_{1} & b_{2} &b_{3} \\ c_{1} & c_{2} &c_{3} \end{vmatrix}\).
Calculation:
Given: \(\rm \vec{a}= \vec i, \vec b = \hat j + \hat k \ and \ \vec c = \hat i + \hat k\)
As we know that scalar triple product \(\rm \vec{a}.( \vec{b}\times \vec{c})= \begin{vmatrix} a_{1} &a_{2} &a_{3}\\ b_{1} & b_{2} &b_{3} \\ c_{1} & c_{2} &c_{3} \end{vmatrix}\)
⇒ \(\rm \vec{a}.( \vec{b}\times \vec{c})= \begin{vmatrix} 1 &0 &0\\ 0 & 1 &1 \\ 1 & 0 &1 \end{vmatrix}\)
⇒ \(\rm \vec{a}.( \vec{b}\times \vec{c})= 1(1-0)-0+0 = 1\)
Hence, option 1 is correct.