Question

Prove property$\left(3\right)$

Short answer

We had proved$u\times v=-(v\times u)$

Step-by-step solution

Step 1. Assumption Of Vectors

Let us assume that the two vectors $u,v$be:

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

Step 2. Solve Left Hand Side

Now, we will solve the lest hand side of the equation,

$\mathit{u}\times \mathit{v}=\left|\begin{array}{lll}\mathit{i}& \mathit{j}& \mathit{k}\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$

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

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

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

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

Step 3. Find the Determinant Form

The above equation can be written in determinant form as,

$=-\left[\begin{array}{ll}{b}_2& c_2\\ b_1& c_1\end{array}|\mathit{i}-|\begin{array}{ll}{a}_2& c_2\\ a_1& c_1\end{array}|\mathit{j}+|\begin{array}{cc}{a}_2& b_2\\ a_1& b_1\end{array}\mid \mathit{k}\right]$

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

$=-(\mathit{v}\times \mathit{u})$

Hence Proved.