Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $\vec{a} = a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$,$\vec{b} = b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$,and $\vec{c} = c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$. Then show that $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$.

(N/A) Given $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$,$\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$,and $\vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$.
First,calculate $(\vec{b}+\vec{c}) = (b_{1}+c_{1}) \hat{i} + (b_{2}+c_{2}) \hat{j} + (b_{3}+c_{3}) \hat{k}$.
Now,$\vec{a} \times(\vec{b}+\vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3} \end{vmatrix}$
$= \hat{i}[a_{2}(b_{3}+c_{3}) - a_{3}(b_{2}+c_{2})] - \hat{j}[a_{1}(b_{3}+c_{3}) - a_{3}(b_{1}+c_{1})] + \hat{k}[a_{1}(b_{2}+c_{2}) - a_{2}(b_{1}+c_{1})]$
$= \hat{i}[a_{2}b_{3} + a_{2}c_{3} - a_{3}b_{2} - a_{3}c_{2}] + \hat{j}[a_{3}b_{1} + a_{3}c_{1} - a_{1}b_{3} - a_{1}c_{3}] + \hat{k}[a_{1}b_{2} + a_{1}c_{2} - a_{2}b_{1} - a_{2}c_{1}] \dots (1)$
Next,$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{vmatrix} = \hat{i}(a_{2}b_{3}-a_{3}b_{2}) + \hat{j}(a_{3}b_{1}-a_{1}b_{3}) + \hat{k}(a_{1}b_{2}-a_{2}b_{1}) \dots (2)$
And $\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{vmatrix} = \hat{i}(a_{2}c_{3}-a_{3}c_{2}) + \hat{j}(a_{3}c_{1}-a_{1}c_{3}) + \hat{k}(a_{1}c_{2}-a_{2}c_{1}) \dots (3)$
Adding $(2)$ and $(3)$ gives:
$(\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) = \hat{i}(a_{2}b_{3} + a_{2}c_{3} - a_{3}b_{2} - a_{3}c_{2}) + \hat{j}(a_{3}b_{1} + a_{3}c_{1} - a_{1}b_{3} - a_{1}c_{3}) + \hat{k}(a_{1}b_{2} + a_{1}c_{2} - a_{2}b_{1} - a_{2}c_{1}) \dots (4)$
Comparing $(1)$ and $(4)$,we see that $\vec{a} \times(\vec{b}+\vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$. Hence proved.