Question

Prove property ($5$).

Short answer

We had proved$u\times (v+w)=(u\times v)+(u\times w)$

Step-by-step solution

Step 1. Assumption Of Vectors

Let the three vectors be:

$$\begin{gathered} \mathit{u}=a_1\mathit{i}+b_1\mathit{j}+c_1\mathit{k}, \\ \mathit{v}=a_2\mathit{i}+b_2\mathit{j}+c_2\mathit{k}\text{and} \\ \mathit{w}=a_3\mathit{i}+b_3\mathit{j}+c_3\mathit{k} \end{gathered}$$

Step 2. Calculating One Side Of the equation

The left hand side of the equation is

$\mathit{u}\times (\mathit{v}+\mathit{w})=\mathit{u}\times \left(\left(a_2\mathit{i}+b_2\mathit{j}+c_2\mathit{k}\right)+\left(a_3\mathit{i}+b_3\mathit{j}+c_3\mathit{k}\right)\right)$

$=\mathit{u}\times \left[\left(a_2+a_3\right)\mathit{i}+\left(b_2+b_3\right)\mathit{j}+\left(c_2+c_3\right)\mathit{k}\right]$

$=\left(a_1\mathit{i}+b_1\mathit{j}+c_1\mathit{k}\right)\times \left[\left(a_2+a_3\right)\mathit{i}+\left(b_2+b_3\right)\mathit{j}+\left(c_2+c_3\right)\mathit{k}\right]$

$=\left|\begin{array}{ccc}\mathit{i}& \mathit{j}& \mathit{k}\\ a_1& b_1& c_1\\ a_2+a_3& b_2+b_3& c_2+c_3\end{array}\right|$

Step 3. Value of the Determinant

On Solving the determinant, we get,

$=\left|\begin{array}{cc}{b}_1& c_1\\ b_2+b_3& c_2+c_3\end{array}\right|\mathit{i}-\left|\begin{array}{cc}{a}_1& c_1\\ a_2+a_3& c_2+c_3\end{array}\right|\mathit{j}+\left|\begin{array}{cc}{a}_1& b_1\\ a_2+a_3& b_2+b_3\end{array}\right|\mathit{k}$

$=\left(b_1{c}_2+b_1{c}_3-b_2{c}_1-b_3{c}_1\right)\mathit{i}-\left(a_1{c}_2+a_1{c}_3-a_2{c}_1-a_3{c}_1\right)\mathit{j}+\left(a_1{b}_2+a_1{b}_3-a_2{b}_1-a_3{b}_1\right)\mathit{k}$

$=\left(b_1{c}_2-b_2{c}_1\right)i+\left(b_1{c}_3-b_3{c}_1\right)\mathit{i}-\left(a_1{c}_2-a_2{c}_1\right)\mathit{j}-\left(a_1{c}_3-a_3{c}_1\right)\mathit{j}+\left(a_1{b}_2-a_2{b}_1\right)\mathit{k}+\left(a_1{b}_3-a_3{b}_1\right)\mathit{k}$

$=\left(b_1{c}_2-b_2{c}_1\right)\mathit{i}-\left(a_1{c}_2-a_2{c}_1\right)\mathit{j}+\left(a_1{b}_2-a_2{b}_1\right)\mathit{k}+\left(b_1{c}_3-b_3{c}_1\right)\mathit{i}-\left(a_1{c}_3-a_3{c}_1\right)\mathit{j}+\left(a_1{b}_3-a_3{b}_1\right)\mathit{k}$

$=\left(\left|\begin{array}{ll}{b}_1& c_1\\ b_2& c_2\end{array}\right|\mathit{i}-\left|\begin{array}{ll}{a}_1& c_1\\ a_2& c_2\end{array}\right|\mathit{j}+\left|\begin{array}{ll}{a}_1& b_1\\ a_2& b_2\end{array}\right|\mathit{k}\right)+\left(\left|\begin{array}{ll}{b}_1& c_1\\ b_3& c_3\end{array}\right|\mathit{i}-\left|\begin{array}{ll}{a}_1& c_1\\ a_3& c_3\end{array}\right|\mathit{j}+\left|\begin{array}{ll}{a}_1& b_1\\ a_3& b_3\end{array}\right|\mathit{k}\right)$

$=\left|\begin{array}{lll}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|+\left|\begin{array}{lll}i& j& k\\ a_1& b_1& c_1\\ a_3& b_3& c_3\end{array}\right|$

$=(u\times v)+(u\times w)$

Hence proved.