Explanation

(a)
Let $\overrightarrow{\mathbf{a}} = a_{1}\widehat{\dot{\mathbf{i}}} + a_{2}\widehat{\dot{\mathbf{j}}} + a_{3}\widehat{\mathbf{k}}$
$and\ \overrightarrow{\mathbf{b}} = b_{1}\widehat{\dot{\mathbf{i}}} + {b\ }_{2}\widehat{\dot{\mathbf{j}}} + b_{3}\widehat{\mathbf{k}}$
$Now,\ \left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \widehat{\dot{\mathbf{i}}} \right\rbrack = \left| \begin{matrix}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
1 & 0 & 0
\end{matrix} \right|$
$= a_{1}(0 - 0) - a_{2}\left( 0 - b_{3} \right) + a_{3}\left( 0 - b_{2} \right) = a_{2}b_{3} - a_{3}b_{2}$
$\therefore 2\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \widehat{\dot{\mathbf{i}}} \right\rbrack\widehat{\dot{\mathbf{i}}} = 2\left\lbrack a_{2}b_{3} - a_{3}b_{2} \right\rbrack\widehat{\dot{\mathbf{i}}}$
Similarly, $2\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \widehat{\dot{\mathbf{j}}} \right\rbrack\widehat{\dot{\mathbf{j}}} = 2\left\lbrack a_{3}b_{1} - a_{1}b_{3} \right\rbrack\widehat{\dot{\mathbf{j}}}$
$and\ 2\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \widehat{\mathbf{k}} \right\rbrack\widehat{\mathbf{k}} = 2\left\lbrack a_{1}b_{2} - a_{2}b_{1} \right\rbrack\widehat{\mathbf{k}}$
$\therefore 2\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \widehat{\dot{\mathbf{i}}} \right\rbrack\widehat{\dot{\mathbf{i}}} + 2\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \widehat{\dot{\mathbf{j}}} \right\rbrack\widehat{\dot{\mathbf{j}}} + 2\left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \widehat{\mathbf{k}} \right\rbrack\widehat{\mathbf{k}} + \left\lbrack \overrightarrow{\mathbf{a}}\ \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{a}}\ \right\rbrack$
$= 2\left\lbrack \left( a_{2}b_{3} - a_{3}b_{2} \right)\widehat{\dot{\mathbf{i}}} + \left( a_{3}b_{1} - a_{1}b_{3} \right)\widehat{\dot{\mathbf{j}}} + \left( a_{1}b_{2} - a_{2}b_{1} \right)\widehat{\mathbf{k}} \right\rbrack$
$= (\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})$