91Ó°ÊÓ

The vector product, and the scalar product of two vectors with $\mathit{θ}$angle between them and unit vector$\hat{\mathbf{n}}$perpendicular to the plane containing them, is defined as:

Vector product:$\stackrel{\mathbf{→}}{\mathbf{A}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\mathbf{\left|}\stackrel{\mathbf{→}}{\mathbf{A}}\mathbf{\right|}\mathbf{\left|}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{\right|}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\hat{\mathbf{n}}$.

Scalar product:$\stackrel{\mathbf{→}}{\mathbf{A}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}\mathbf{\left|}\stackrel{\mathbf{→}}{\mathbf{A}}\mathbf{\right|}\mathbf{\left|}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{\right|}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}$.

For two vectors $\stackrel{→}{A}$and $\stackrel{→}{B}$,

$$\begin{gathered} (\stackrel{→}{A}×\stackrel{→}{B})=\left|\stackrel{→}{A}\right|\left|\stackrel{→}{B}\right|sin\mathrm{θ}\hat{n} \\ (\stackrel{→}{A}×\stackrel{→}{B}{)}^2=\left(\right|\stackrel{→}{A}\left|\right|\stackrel{→}{B}|sin\mathrm{θ}\hat{n}{)}^2 \\ =\left|\stackrel{→}{A}{|}^2\right|\stackrel{→}{B}{|}^{2}si{n}^2\mathrm{θ} \end{gathered}$$

Therefore $(\stackrel{→}{A}×\stackrel{→}{B}{)}^2=A^2{B}^{2}si{n}^2\mathrm{θ}…\left(1\right)$

$Q\left|\stackrel{→}{A}\right|=A,\left|\stackrel{→}{B}\right|=B,(\hat{n}{)}^2=\hat{n}×\hat{n}=1$

$$\begin{gathered} (\stackrel{→}{A}×\stackrel{→}{B})=\left|\stackrel{→}{A}\right|\left|\stackrel{→}{B}\right|cos\mathrm{θ} \\ (\stackrel{→}{A}×\stackrel{→}{B}{)}^2=\left(\right|\stackrel{→}{A}\left|\right|\stackrel{→}{B}|cos\mathrm{θ}\hat{n}{)}^2 \\ =\left|\stackrel{→}{A}{|}^2\right|\stackrel{→}{B}{|}^{2}co{s}^2\mathrm{θ} \\ (\stackrel{→}{A}×\stackrel{→}{B}{)}^2=A^2{B}^{2}co{s}^2\mathrm{θ}…\left(2\right) \end{gathered}$$

From results (1) and (2)

$$\begin{gathered} (\stackrel{→}{A}×\stackrel{→}{B}{)}^2+(\stackrel{→}{A}×\stackrel{→}{B}{)}^2=\left|\stackrel{→}{A}{|}^2\right|\stackrel{→}{B}{|}^{2}si{n}^2\mathrm{θ}+\left|\stackrel{→}{A}{|}^2\right|\stackrel{→}{B}{|}^{2}co{s}^2\mathrm{θ} \\ =\left|\stackrel{→}{A}{|}^2\right|\stackrel{→}{B}{|}^2 \\ =A^2{B}^2 \end{gathered}$$

Thus$(\stackrel{→}{A}×\stackrel{→}{B}{)}^2+(\stackrel{→}{A}×\stackrel{→}{B}{)}^2=\left|\stackrel{→}{A}{|}^2\right|\stackrel{→}{B}{|}^2=A^2{B}^2$

Hence, the required value is $(\stackrel{→}{A}×\stackrel{→}{B}{)}^2+(\stackrel{→}{A}×\stackrel{→}{B}{)}^2=\left|\stackrel{→}{A}{|}^2\right|\stackrel{→}{B}{|}^2=A^2{B}^2$.