Chapter 9, Problem 23RE

(a)

To determine

To find:The component of $u$ along $v$ .

Explanation of Solution

Given: $u=i+2j$ and $v=4i-9j$

The component of $u$ along $v$ is calculated as,

  $\begin{array}{c}{\text{comp}}_{v}u=\frac{u\cdot v}{\left|v\right|}\\ =\frac{\left(1\times 4\right)+\left(2\times \left(-9\right)\right)}{\sqrt{4^2+{\left(-9\right)}^2}}\\ =\frac{-14}{\sqrt{97}}\end{array}$

Therefore, the component of $u$ along $v$ is $\frac{-14}{\sqrt{97}}$ .

(b)

To determine

To find :The ${\text{proj}}_{v}u$ .

Explanation of Solution

The ${\text{proj}}_{v}u$ is calculated as,

  $\begin{array}{c}{\text{proj}}_{v}u=\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v\\ =\frac{-14}{{\left(\sqrt{97}\right)}^2}\left(4i-9j\right)\\ =\frac{-56}{97}i+\frac{126}{97}j\end{array}$

Therefore, the ${\text{proj}}_{v}u$ is $\frac{-56}{97}i+\frac{126}{97}j$ .

(c)

To determine

To resolve :The vector $u$ into $u_1$ and $u_2$ .

Explanation of Solution

The component of $u$ parallel to $v$ is calculated as,

  $\begin{array}{c}{u}_1={\text{proj}}_{v}u\\ =\left(\frac{u\cdot v}{{\left|v\right|}^2}\right)v\\ =\frac{-14}{{\left(\sqrt{97}\right)}^2}\left(4i-9j\right)\\ =\frac{-56}{97}i+\frac{126}{97}j\end{array}$

The component of $u$ perpendicular to $v$ is calculated as,

  $\begin{array}{c}{u}_2=u-u_1\\ =\left(i+2j\right)-\left(\frac{-56}{97}i+\frac{126}{97}j\right)\\ =\frac{153}{97}i+\frac{68}{97}j\end{array}$

Therefore, the component of $u$ parallel to $v$ is calculatedis $\frac{-56}{97}i+\frac{126}{97}j$ , the component of $u$ perpendicular to $v$ is calculated is $\frac{153}{97}i+\frac{68}{97}j$ .