Question

Show that $\mathit{B}\mathbf{\left|}\mathit{A}\mathbf{\right|}\mathbf{}\mathbf{+}\mathbf{}\mathit{A}\mathbf{\left|}\mathit{B}\mathbf{\right|}$and$\mathit{A}\mathbf{\left|}\mathit{B}\mathbf{\right|}\mathbf{}\mathbf{-}\mathbf{}\mathit{B}\mathbf{\left|}\mathit{A}\mathbf{\right|}$are orthogonal.

Short answer

$B\left|A\right|+A\left|B\right|$and$A\left|B\right|-B\left|A\right|$ are orthogonal.

Step-by-step solution

Concept and formula used for the given vector

Write the expression for perpendicular vectors.

$\mathit{P}\mathbf{\times }\mathit{Q}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, $\mathit{P}$and $\mathit{Q}$are vectors.

To show that the given vectors are orthogonal

To calculate the dot product of the vectors,

$$\begin{gathered} \Rightarrow P·Q \\ \Rightarrow \left(B\right|A|+A|B\left|\right)·\left(A\right|B|-B|A\left|\right) \\ \Rightarrow \left(A\right|B|{)}^2-\left(B\right|A|{)}^2 \\ \Rightarrow (A·A{)}^2\left(\right|B|{)}^2-(B·B{)}^2\left(\right|A|{)}^2 \\ \Rightarrow \left(\right|A|{)}^2\left(\right|B|{)}^2-\left(\right|B|{)}^2\left(\right|A|{)}^2 \\ \Rightarrow 0 \end{gathered}$$

Therefore, the scalar product of vectors is zero. Thus, the vectors are perpendicular.