Question

What is the value of $\mathbf{(}\mathbf{A}\mathbf{\times }\mathbf{B}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{A}\mathbf{\times }\mathbf{B}{\mathbf{)}}^{\mathbf{2}}\mathbf{?}$Comment: This is a special case of Lagrange's identity. (See Chapter 6. Problem 3.12b, page 284.)

Use vectors as in Problems 3 to 8, and also the dot and cross product, to prove the following theorems from geometry.

Short answer

The required value is$(\overrightarrow{A}\times \overrightarrow{B}{)}^2+(\overrightarrow{A}\times \overrightarrow{B}{)}^2=\left|\overrightarrow{A}{|}^2\right|\overrightarrow{B}{|}^2=A^2{B}^2$

Step-by-step solution

Concept and formula used

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

Vector product:$\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{\left|}\overrightarrow{\mathbf{A}}\mathbf{\right|}\mathbf{\left|}\overrightarrow{\mathbf{B}}\mathbf{\right|}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\hat{\mathbf{n}}$.

Scalar product:$\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{\left|}\overrightarrow{\mathbf{A}}\mathbf{\right|}\mathbf{\left|}\overrightarrow{\mathbf{B}}\mathbf{\right|}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$.

To find the value of the given vector

For two vectors $\overrightarrow{A}$and $\overrightarrow{B}$,

$$\begin{gathered} (\overrightarrow{A}\times \overrightarrow{B})=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|sin\theta \hat{n} \\ (\overrightarrow{A}\times \overrightarrow{B}{)}^2=\left(\right|\overrightarrow{A}\left|\right|\overrightarrow{B}|sin\theta \hat{n}{)}^2 \\ =\left|\overrightarrow{A}{|}^2\right|\overrightarrow{B}{|}^{2}si{n}^2\theta \end{gathered}$$

Therefore $(\overrightarrow{A}\times \overrightarrow{B}{)}^2=A^2{B}^{2}si{n}^2\theta \dots \left(1\right)$

$Q\left|\overrightarrow{A}\right|=A,\left|\overrightarrow{B}\right|=B,(\hat{n}{)}^2=\hat{n}\times \hat{n}=1$

$$\begin{gathered} (\overrightarrow{A}\times \overrightarrow{B})=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|cos\theta \\ (\overrightarrow{A}\times \overrightarrow{B}{)}^2=\left(\right|\overrightarrow{A}\left|\right|\overrightarrow{B}|cos\theta \hat{n}{)}^2 \\ =\left|\overrightarrow{A}{|}^2\right|\overrightarrow{B}{|}^{2}co{s}^2\theta \\ (\overrightarrow{A}\times \overrightarrow{B}{)}^2=A^2{B}^{2}co{s}^2\theta \dots \left(2\right) \end{gathered}$$

From results (1) and (2)

$$\begin{gathered} (\overrightarrow{A}\times \overrightarrow{B}{)}^2+(\overrightarrow{A}\times \overrightarrow{B}{)}^2=\left|\overrightarrow{A}{|}^2\right|\overrightarrow{B}{|}^{2}si{n}^2\theta +\left|\overrightarrow{A}{|}^2\right|\overrightarrow{B}{|}^{2}co{s}^2\theta \\ =\left|\overrightarrow{A}{|}^2\right|\overrightarrow{B}{|}^2 \\ =A^2{B}^2 \end{gathered}$$

Thus$(\overrightarrow{A}\times \overrightarrow{B}{)}^2+(\overrightarrow{A}\times \overrightarrow{B}{)}^2=\left|\overrightarrow{A}{|}^2\right|\overrightarrow{B}{|}^2=A^2{B}^2$

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