Let the vectors $\vec a,\vec b,\vec c$ be given as ${a_1}\hat i + {a_2}\hat j + {a_3}\hat k,{b_1}\hat i + {b_2}\hat j + {b_3}\hat k,{c_1}\hat i + {c_2}\hat j + {c_3}\hat k$. Then show that $\vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c$.

$\vec b + \vec c = ({b_1}\hat i + {b_2}\hat j + {b_3}\hat k) + ({c_1}\hat i + {c_2}\hat j + {c_3}\hat k)$

$= ({b_1} + {c_1})\hat i + ({b_2} + {c_2})\hat j + ({b_3} + {c_3})\hat k$

$\therefore$ L.H.S. $= \vec a \times (\vec b + \vec c) = \left| {\begin{array}{cccccccccccccccccccc}{\hat i}&{\hat j}&{\hat k}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1} + {c_1}}&{{b_2} + {c_2}}&{{b_3} + {c_3}}\end{array}} \right|$.

$= [{a_2}({b_3} + {c_3}) - {a_3}({b_2} + {c_2})]\hat i - [{a_1}({b_3} + {c_3}) - {a_3}({b_1} + {c_1})]\hat j + [{a_1}({b_2} + {c_2}) - {a_2}({b_1} + {c_1})]\hat k$ $= [({a_2}{b_3} - {a_3}{b_2}) + ({a_2}{c_3} - {a_3}{c_2})]\hat i - [({a_1}{b_3} - {a_3}{b_1}) + ({a_1}{c_3} - {a_3}{c_1})]\hat j$ $+ [({a_1}{b_2} - {a_2}{b_1}) + ({a_1}{c_2} - {a_2}{c_1})]\hat k$

…(i)
R.H.S. $= \vec a \times \vec b + \vec a \times \vec c = \left| {\begin{array}{llllllllllllllllllll}{\hat i}&{\hat j}&{\hat k}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right| + \left| {\begin{array}{llllllllllllllllllll}{\hat i}&{\hat j}&{\hat k}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|$

$= [({a_2}{b_3} - {a_3}{b_2})\hat i - ({a_1}{b_3} - {a_3}{b_1})\hat j + ({a_1}{b_2} - {a_2}{b_1})\hat k]$ $+ [({a_2}{c_3} - {a_3}{c_2})\hat i - ({a_1}{c_3} - {a_3}{c_1})\hat j + ({a_1}{c_2} - {a_2}{c_1})\hat k]$

…(ii)
From (i) and (ii),

we get $\vec a \times (\vec b + \vec c) = \vec a \times \vec b + \vec a \times \vec c$