Question

Let the vectors $\vec{a},\vec{b},\vec{c}$ 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 tha$t=\vec{a}\times (\vec{b}+\vec{c})=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$

Answer 1

We have,
$\vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k},\vec{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k},\vec{c}=c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}$
$(\vec{b}+\vec{c})=(b_{1}+c_{1})\hat{i}+(b_{2}+c_{2})\hat{j}+((b_{3}+c_{3})\hat{k}$
Now,$\vec{a}\times (\vec{b}+\vec{c})\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ a_{1} & a_2 & a_3\\\b_1+c_1&b_2+c_2&b_3+c_3\end{vmatrix}$
$=\hat{i}$[a2(b2+c3)-a3(b2+c2)]-$=\hat{j}$[a1(b3+c3)-a3(b1+c1)+$\hat{k}$[a1(b2+c2)-a2(b1+c1)]
$=\hat{i}$[a2b3+a2c3-a3b2-a3c2]+$\hat{j}$[-a1b3-a1c3+a3b1+a3c1]+$\hat{k}$[a1b2+a1c2-a2b1-a2c1]...(1)
$\vec{a}\times\vec{b}$=$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ a_1 & a_2 & a_3 \\\b_1&b_2&b_3\end{vmatrix}$
$=\hat{i}$[a2b3-a3b2]+$\hat{j}$[b1a3-a1b3]+$\hat{k}$[a1b2-a2b1]...(2)
$\vec{a}\times\vec{c}$=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ a_1 & a_2 & a_3 \\\c_1&c_2&c_3\end{vmatrix}
=$\hat{i}$[a2c3-a3c2]+$\hat{j}$[a3c1-a1c3]$+\hat{k}$[a1c2-a2c1]...(3)
On adding (2)and(3),we get:
($\vec{a}\times\vec{b}$)+($\vec{a}\times\vec{c}$)$=\hat{i}$[a2b3+a2c3-a3b2-a3c2]+$\hat{j}$[b1a3+a3c1-a1b3-a1c3]+$\hat{k}$[a1b2+a1c2-a2b1-a2c1]...(4)
Now,from (1)and(4),we have:
$\vec{a}\times (\vec{b}+\vec{c})$=$\vec{a}\times\vec{b}+\vec{a}\times\vec{c}$
Hence,the given result is proved.