Chapter 6, Problem 18CT

To determine

To find: The projection of the vectors $u$ on $v$ and then write the sum of the orthogonal vectors one of which is ${\text{proj}}_{v}u$ .

Answer to Problem 18CT

The value of the orthogonal vector is $\langle \frac{185}{26},\frac{37}{26}\rangle +\langle \frac{-29}{26},\frac{145}{26}\rangle$ and $u$ as the sum of the orthogonal vector is $\langle \frac{185}{26},\frac{37}{26}\rangle +\langle \frac{-29}{26},\frac{145}{26}\rangle$ .

Explanation of Solution

Given:

The value of given vectors is $u=\langle 6,7\rangle$ and $v=\langle -5,-1\rangle$ .

Calculation:

Consider the given vector is,

  $\begin{array}{l}u=\langle 6,7\rangle \\ v=\langle -5,-1\rangle \end{array}$

Consider the $u$ and $v$ are the two non-zero vectors and the projection of vector $u$ on $v$ is,

  $\begin{array}{c}{\text{proj}}_{v}u=\left(\frac{u\cdot v}{\left|\left|v\right|\right|}\right)v\\ =\left(\frac{\langle 6,7\rangle \cdot \langle -5,-1\rangle }{\left|\left|\sqrt{{\left(-5\right)}^2+{\left(-1\right)}^2}\right|\right|}\right)\langle -5,-1\rangle \\ =\frac{-37}{26}\langle -5.-1\rangle \end{array}$

Consider the $w_1$ as the orthogonal vector, then,

  $\begin{array}{c}{w}_1={\text{proj}}_{v}u\\ =\left(\frac{u\cdot v}{{\left|\left|v\right|\right|}^2}\right)v\\ =\frac{-37}{26}\langle -5,-1\rangle \end{array}$

Consider the value of the orthogonal vectors $w_2$ as,

  $w_2=u-w_1$

Then,

  $\begin{array}{c}{w}_2=\langle 6,7\rangle -\frac{-37}{26}\langle -5,-1\rangle \\ =\langle \frac{-29}{26},\frac{145}{26}\rangle \end{array}$

The value of $u$ as the sum of two orthogonal vector is,

  $u=w_1+w_2$

Then,

  $\begin{array}{c}u=w_1+w_2\\ =\langle \frac{185}{26},\frac{37}{26}\rangle +\langle \frac{-29}{26},\frac{145}{26}\rangle \end{array}$