Chapter 9.5, Problem 40AYU

To determine

To prove: the given vector $\text{v}-\alpha \text{w}$ is orthogonal if $\alpha =\frac{\text{v}\cdot \text{w}}{{\Vert \text{w}\Vert }^2}$ .

Explanation of Solution

Given:

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

  $\alpha =\frac{\text{v}\cdot \text{w}}{{\Vert \text{w}\Vert }^2}$

Calculation:

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

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

It is given that $\alpha =\frac{\text{v}\cdot \text{w}}{{\Vert \text{w}\Vert }^2}$ .

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

Therefore,

  $\begin{array}{l}\left(\text{v}-\alpha \text{w}\right)\cdot \text{w}\\ =\left(a_{1}i+b_{2}j-\frac{\left(a_{1}i+b_{2}j\right)\cdot \left(a_{1}i+b_{2}j\right)}{{\Vert \text{w}\Vert }^2}\right)\cdot \left(a_{1}i+b_{2}j\right)\\ =\left(a_{1}i+b_{2}j-\frac{\left(a_{1}i+b_{2}j\right)\cdot \left(a_{1}i+b_{2}j\right)}{\left(a_2^2+b_2^2\right)}\right)\cdot \left(a_{1}i+b_{2}j\right)\\ =\left(-\frac{\left(a_1{b}_2^2+a_2{b}_1{b}_2\right)a_2+\left(b_1{a}_2^2-a_2{a}_1{b}_2\right)b_2}{\left(a_2^2+b_2^2\right)}\right)\\ =0\end{array}$

Hence, the given vector $\text{v}-\alpha \text{w}$ is orthogonal $\alpha =\frac{\text{v}\cdot \text{w}}{{\Vert \text{w}\Vert }^2}$ .