Prove for vectors u and vthat

${\left|\left|\mathit{u}\times \mathit{v}\right|\right|}^2={\left|\left|\mathit{u}\right|\right|}^2{\left|\left|\mathit{v}\right|\right|}^2-{\left(\mathit{u}\mathbf{·}\mathit{v}\right)}^2$

Consider the given question,

${\left|\left|\mathit{u}\times \mathit{v}\right|\right|}^2={\left|\left|\mathit{u}\right|\right|}^2{\left|\left|\mathit{v}\right|\right|}^2-{\left(\mathit{u}\mathbf{·}\mathit{v}\right)}^2......\left(i\right)$

Assume $$\begin{gathered} \mathit{u}=a_1\mathit{i}+b_1\mathit{j}+c_1\mathit{k}, \\ u=a_2\mathit{i}+b_2\mathit{j}+c_2\mathit{k} \end{gathered}$$.

Cross product is given below,

$\mathit{v}\times \mathit{u}=\left(b_1{c}_2-b_2{c}_1\right)\mathit{i}-\left({\mathrm{a}}_1{\mathrm{c}}_2-{\mathrm{a}}_2{\mathrm{c}}_1\right)\mathbf{j}+\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)\mathbf{k}$

Area denoted as,

$$\begin{gathered} \left|\left|\mathit{v}\mathbf{\times }\mathit{u}\right|\right|=\left|\left|\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1\right)\mathbf{i}-\left({\mathrm{a}}_1{\mathrm{c}}_2-{\mathrm{a}}_2{\mathrm{c}}_1\right)\mathbf{j}+\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)\mathbf{k}\right|\right| \\ =\sqrt{{\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1\right)}^2-{\left({\mathrm{a}}_1{\mathrm{c}}_2-{\mathrm{a}}_2{\mathrm{c}}_1\right)}^2+{\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)}^2} \end{gathered}$$

Squaring both the sides,

$$\begin{gathered} {\left|\left|\mathit{v}\mathbf{\times }\mathit{u}\right|\right|}^2={\left(\sqrt{{\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1\right)}^2-{\left({\mathrm{a}}_1{\mathrm{c}}_2-{\mathrm{a}}_2{\mathrm{c}}_1\right)}^2+{\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)}^2}\right)}^2 \\ ={b_1}^2{c_2}^2-2b_1{c}_2{b}_2{c}_1+{b_2}^2{c_1}^2+{a_1}^2{c_2}^2-2a_1{c}_2{a}_2{c}_1+{a_2}^2{c_1}^2+{a_1}^2{b_2}^2-2a_1{b}_2{a}_2{b}_1+{a_2}^2{b_1}^2 \end{gathered}$$

Take RHS of equation (i),

$$\begin{gathered} \left|\left|\mathit{u}\right|\right|=\left|\left|a_1\mathit{i}+b_1\mathit{j}+c_1\mathit{k}\right|\right| \\ \left|\left|\mathit{u}\right|\right|=\sqrt{{a_1}^2+{b_1}^2+{c_1}^2} \\ {\left|\left|\mathit{u}\right|\right|}^2={a_1}^2+{b_1}^2+{c_1}^2 \\ \left|\left|\mathit{v}\right|\right|=\left|\left|a_2\mathit{i}+b_2\mathit{j}+c_2\mathit{k}\right|\right| \\ \left|\left|\mathit{v}\right|\right|=\sqrt{{a_2}^2+{b_2}^2+{c_2}^2} \\ {\left|\left|\mathit{v}\right|\right|}^2={a_2}^2+{b_2}^2+{c_2}^2 \end{gathered}$$

Take the dot product,

$$\begin{gathered} \left(\mathit{u}\mathbf{·}\mathit{v}\right)=\left(a_1\mathit{i}+b_1\mathit{j}+c_1\mathit{k}\right)·\left(a_2\mathit{i}+b_2\mathit{j}+c_2\mathit{k}\right) \\ \left(\mathit{u}\mathbf{·}\mathit{v}\right)=a_1{a}_2+b_1{b}_2+c_1{c}_2 \\ {\left(\mathit{u}\mathbf{·}\mathit{v}\right)}^2={\left(a_1{a}_2+b_1{b}_2+c_1{c}_2\right)}^2 \end{gathered}$$

Then,

$$\begin{gathered} {\left|\left|\mathit{u}\right|\right|}^2{\left|\left|\mathit{v}\right|\right|}^2-{\left(\mathit{u}\mathbf{·}\mathit{v}\right)}^2=\left({a_1}^2+{b_1}^2+{c_1}^2\right)\left({a_2}^2+{b_2}^2+{c_2}^2\right)-{\left(a_1{a}_2+b_1{b}_2+c_1{c}_2\right)}^2 \\ ={a_1}^2{b_2}^2+{a_1}^2{c_2}^2+{a_2}^2{b_1}^2+{b_1}^2{c_1}^2+{b_1}^2{c_1}^2+{b_2}^2{c_1}^2-2a_1{a}_2{b}_1{b}_2-2a_1{a}_2{c}_1{c}_2-2b_1{b}_2{c}_1{c}_2 \end{gathered}$$

Therefore,$LHS=RHS$.

Hence, ${\left|\left|\mathit{u}\times \mathit{v}\right|\right|}^2={\left|\left|\mathit{u}\right|\right|}^2\left|\left|\mathit{v}\right|\right|-{\left(\mathit{u}\mathbf{·}\mathit{v}\right)}^2$proved.