Chapter 6, Problem 84CR

To determine

To calculate:

The project of vector $u$ onto $v$ and write vector $u$ as the sum of any two orthogonal vectors:

  $\begin{array}{l}u=\langle 5,6\rangle \\ v=\langle 10,0\rangle \end{array}$

Answer to Problem 84CR

The project of vector $u$ onto $v$ is $pro{j}_{v}u=\left(5,0\right)$ .

Explanation of Solution

Given information:

  $\begin{array}{l}u=\langle 5,6\rangle \\ v=\langle 10,0\rangle \end{array}$

Formula used:

The project of vector $u$ onto $v$ is given by the following expression:

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

Calculation:

Know that:

  $\begin{array}{l}\langle u_1,u_2\rangle =\langle 5,6\rangle \\ \langle v_1,v_2\rangle =\langle 10,0\rangle \end{array}$

The dot product of given two vectors $u$ and $v$ is calculated as:

  $\begin{array}{l}u\cdot v=u_1{v}_1+u_2{v}_2\\ u\cdot v=\left(5\right)\left(10\right)+\left(6\right)\left(0\right)\\ u\cdot v=50-0\\ u\cdot v=50\end{array}$

Now, the magnitude of vector $u$ and $v$ is calculated as:

  $\Vert v\Vert =\sqrt{v_1{}^2+v_2{}^2}=\sqrt{{\left(10\right)}^2+{\left(0\right)}^2}=10$

The project of vector $u$ onto $v$ is calculated as:

  $\begin{array}{l}pro{j}_{v}u=\left(\frac{u\cdot v}{{\Vert v\Vert }^2}\right)v\\ pro{j}_{v}u=\left(\frac{50}{{\left(10\right)}^2}\right)\left(10,0\right)\\ pro{j}_{v}u=\left(0.5\right)\left(10,0\right)\\ pro{j}_{v}u=\left(5,0\right)\end{array}$

So, this is first orthogonal vector $w_1=\left(5,0\right)$ . And other component is:

  $\begin{array}{l}{w}_2=u-w_1\\ w_2=\left(5,6\right)-\left(5,0\right)\\ w_2=\left(0,6\right)\end{array}$

Now, write vector $u$ as the sum of any two orthogonal vectors $w_1$ and $w_2$ :

  $\begin{array}{l}u=w_1+w_2\\ u=\left(5,0\right)+\left(0,6\right)\\ u=\left(5,6\right)\end{array}$

Hence, vector $u$ as the sum of any two orthogonal vectors one of which is $pro{j}_{v}u$ .