Chapter 6, Problem 86RE

To determine

To find: The projection of the vector and the representation of the vector as the sum of two orthogonal vectors.

Answer to Problem 86RE

The projection of the vector is $\langle 5,0\rangle$ and the representation of the vector $u$ in terms of orthogonal vectors is $u=\langle 5,0\rangle +\langle 0,6\rangle$ .

Explanation of Solution

Given information:

The given vectorsare,

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

Concept used:

The projection of $u$ onto $v$ is calculated as,

  $\begin{array}{c}{w}_1={\text{proj}}_{v}u\\ =\left(\frac{u\cdot v}{{\Vert v\Vert }^2}\right)v\end{array}$

The projection of $u$ onto $v$ is calculated as,

  $\begin{array}{c}{w}_1={\text{proj}}_{v}u\\ =\left(\frac{u\cdot v}{{\Vert v\Vert }^2}\right)v\\ =\left(\frac{5\times 10+\left(6\right)\left(0\right)}{\left({\left(10\right)}^2+{\left(0\right)}^2\right)}\right)\langle 10,0\rangle \\ =\langle 5,0\rangle \end{array}$

The other component $w_2$ is calculated as,

  $\begin{array}{c}{w}_2=u-w_1\\ =\langle 5,6\rangle -\langle 5,0\rangle \\ =\langle 0,6\rangle \end{array}$

The representation of the vector $u$ in terms of orthogonal vectors as,

  $\begin{array}{c}u=w_1+w_2\\ u=\langle 5,0\rangle +\langle 0,6\rangle \end{array}$

Therefore, the projection of the vector is $\langle 5,0\rangle$ and the representation of the vector $u$ in terms of orthogonal vectors is $u=\langle 5,0\rangle +\langle 0,6\rangle$ .