Chapter 6, Problem 19CT

To determine

Find the projection of $u=\langle 6,7\rangle ontov=\langle -5,-1\rangle .$ Then write $u$ as the sum of two orthogonal vectors, one of which is proj,u.

Answer to Problem 19CT

The vector u as a sum of two orthogonal vectors is $\langle \frac{185}{26},\frac{37}{26}\rangle +\langle -\frac{29}{26},\frac{145}{26}\rangle$

Explanation of Solution

Given:

Consider the two vectors,

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

The projection of vector u on the vector v is given by,

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

Therefore, the projection of u on v is,

  $\begin{array}{cc}pro{j}_{v}u& =\left(\frac{\langle 6,7\rangle \cdot \langle -5,-1\rangle }{{(\sqrt{{(-5)}^2+{(-1)}^2})}^2}\right)\langle -5,-1\rangle \\ & =\left(\frac{(6)(-5)+(7)(-1)}{26}\right)\langle -5,-1\rangle \\ & =\left(-\frac{37}{26}\right)\langle -5,-1\rangle \\ & =\langle \frac{185}{26},\frac{37}{26}\rangle \end{array}$

Now the component of u which is perpendicular to the vector projvu is,

  $\begin{array}{cc}{u}_1& =u-pro{j}_{v}u\\ & =\langle 6,7\rangle -\langle \frac{185}{26},\frac{37}{26}\rangle \\ & =\langle 6-\frac{185}{26},7-\frac{37}{26}\rangle \\ & =\langle -\frac{29}{26},\frac{145}{26}\rangle \end{array}$

Therefore, the vector u as a sum of two orthogonal vectors is

  $u=pro{j}_{v}u+u_1$

  $\langle \frac{185}{26},\frac{37}{26}\rangle +\langle -\frac{29}{26},\frac{145}{26}\rangle$

  $$ .