Chapter 9.2, Problem 43E

To determine

To prove: The ${\text{proj}}_{v}u$ and $u{\text{-proj}}_v$ are orthogonal.

Explanation of Solution

Given information:

The ${\text{proj}}_{v}u$ and $u{\text{-proj}}_v$ .

Calculation:

Consider the vector.

  $\begin{array}{l}u=\langle x_1,y_1\rangle \\ v=\langle x_2,y_2\rangle \end{array}$

The value of ${\text{proj}}_{v}u$ is,

  $\begin{array}{c}{\text{proj}}_{v}u=\left(\frac{uv}{{\left|v\right|}^2}\right)v\\ =\left(\frac{x_1{x}_2+y_1{y}_2}{x_2^2+y_2^2}\right)\langle x_2,y_2\rangle \\ =\langle \frac{x_1{x}_2^2+x_2{y}_2{y}_1}{x_2^2+y_2^2},\frac{x_1{x}_2{y}_2-y_1{y}_2^2}{x_2^2+y_2^2}\rangle \end{array}$

The value of ${\text{proj}}_{v}u$ is,

  $\begin{array}{c}u{\text{-proj}}_{v}u=\langle x_1,y_2\rangle -\langle \frac{x_1{x}_2^2+x_2{y}_2{y}_1}{x_2^2+y_2^2},\frac{x_1{x}_2{y}_2-y_1{y}_2^2}{x_2^2+y_2^2}\rangle \\ =\langle \frac{x_1{y}_2^2-x_2{y}_2{y}_1}{x_2^2+y_2^2},\frac{y_1{x}_2^2-x_1{x}_2{y}_2}{x_2^2+y_2^2}\rangle \end{array}$

The dot product is,

  $\begin{array}{c}\left({\text{proj}}_{v}u\right)\cdot \left(u{\text{-proj}}_{v}u\right)=\langle \frac{x_1{y}_2^2-x_2{y}_2{y}_1}{x_2^2+y_2^2},\frac{y_1{x}_2^2-x_1{x}_2{y}_2}{x_2^2+y_2^2}\rangle \cdot \langle \frac{x_1{x}_2^2+x_2{y}_2{y}_1}{x_2^2+y_2^2},\frac{x_1{x}_2{y}_2-y_1{y}_2^2}{x_2^2+y_2^2}\rangle \\ =0\end{array}$

Thus, the ${\text{proj}}_{v}u$ and $u{\text{-proj}}_v$ are orthogonal.