Chapter 6.2, Problem 58E

Solution

a.

To prove: The projection of a vector.

Given information:

The $u$ and $v$ are non-zero vectors.

Proof:

The given identity is,

  $pro{j}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v$

Here, $u,v\ne 0$

Now, draw the vectors $u,v$ and projection of $u$ onto $v$ .

  

The vector $\overrightarrow{PR}$ is given by,

  $\overrightarrow{PR}=\frac{PR}{PS}v$

From the graph, to find the value of $PR$

  $\begin{array}{c}\cos QPS=\frac{PR}{PQ}\\ PR=PQ\cos QPS\\ =\left|u\right|\cdot \frac{u\cdot v}{\left|u\right|\cdot \left|v\right|}\\ =\frac{u\cdot v}{\left|v\right|}\end{array}$

To find $PS$

  $PS=\left|v\right|$

Now to find the projection of $u$ ,

  $\begin{array}{c}pro{j}_{v}u=\overrightarrow{PR}\\ =\frac{PR}{PS}v\\ =\frac{\frac{u\cdot v}{\left|v\right|}}{\left|v\right|}v\\ pro{j}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v\end{array}$

Hence $pro{j}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v$ is proved.

b.

To prove: The projection of a vector.

Given information:

The $u$ and $v$ are non-zero vectors.

Proof:

The given identity is,

  $(u-pro{j}_{v}u).(pro{j}_{v}u)=0$

Here, $u,v\ne 0$

Now, draw the vectors $u,v$ and projection of $u$ onto $v$ .

  

Consider,

  $\begin{array}{c}(u-pro{j}_{v}u).(pro{j}_{v}u)=(u-\overrightarrow{PR})\cdot \overrightarrow{PR}\\ =\overrightarrow{RQ}\cdot \overrightarrow{PR}\end{array}$

Since $RQ$ is perpendicular to $PS$ , the vectors $\overrightarrow{PR}$ and $\overrightarrow{RQ}$ are orthogonal. So their scalar product is zero.

  $\overrightarrow{RQ}\cdot \overrightarrow{PR}=0$

Therefore,

  $(u-pro{j}_{v}u).(pro{j}_{v}u)=0$

Hence $(u-pro{j}_{v}u).(pro{j}_{v}u)=0$ is proved.