Chapter 9.5, Problem 41AYU

To determine

To prove: the given vector $\Vert \text{w}\Vert \text{v}+\Vert \text{v}\Vert \text{w and }\Vert \text{w}\Vert \text{v}-\Vert \text{v}\Vert \text{w }$ are orthogonal.

Explanation of Solution

Given:

The givenvector $\text{v and w}$ denote two non-zero vector.

Calculation:

As per the question,the vector $\text{v and w}$ denote two non-zero vector.

Now, let $v=a_{1}i+b_{1}j\text{ and }w=a_{2}i+b_{2}j$ .

Two vectors are orthogonal if and only if dot product of two vectors is zero.

Therefore,

  $\begin{array}{l}\left(\Vert \text{w}\Vert \text{v}+\Vert \text{v}\Vert \text{w}\right).\left(\Vert \text{w}\Vert \text{v}-\Vert \text{v}\Vert \text{w}\right)\\ =\left[\Vert \text{w}\Vert \left(a_{1}i+b_{1}j\right)+\Vert \text{v}\Vert \left(a_{2}i+b_{2}j\right)\right]\cdot \left[\Vert \text{w}\Vert \left(a_{1}i+b_{1}j\right)-\Vert \text{v}\Vert \left(a_{2}i+b_{2}j\right)\right]\\ =\left[\left(\Vert \text{w}\Vert a_1+\Vert \text{v}\Vert a_2\right)\left(\Vert \text{w}\Vert a_1-\Vert \text{v}\Vert a_2\right)\right]+\left[\left(\Vert \text{w}\Vert b_1+\Vert \text{v}\Vert b_2\right)\left(\Vert \text{w}\Vert b_1-\Vert \text{v}\Vert b_2\right)\right]\\ ={\Vert \text{w}\Vert }^2{a}_1^2-\Vert \text{v}\Vert a_2^2+{\Vert \text{w}\Vert }^2{b}_1^2-\Vert \text{v}\Vert b_2^2\\ ={\Vert \text{w}\Vert }^2\left(a_1^2+b_1^2\right)-\Vert \text{v}\Vert \left(a_2^2+b_2^2\right)\\ =\left(a_2^2+b_2^2\right)\left(a_1^2+b_1^2\right)-\left(a_1^2+b_1^2\right)\left(a_2^2+b_2^2\right)\\ =0\end{array}$

Hence, the given vector $\Vert \text{w}\Vert \text{v}+\Vert \text{v}\Vert \text{w and }\Vert \text{w}\Vert \text{v}-\Vert \text{v}\Vert \text{w }$ are orthogonal.