Chapter 6.3, Problem 50E

To determine

To find the vector v in the direction of the vector $\text{u}=\langle 3,3\rangle$ .

Answer to Problem 50E

  $\text{v}=\langle 4\sqrt{2},4\sqrt{2}\rangle$

Explanation of Solution

Given:

  $\text{u}=\langle 3,3\rangle \&\Vert \text{v}\Vert =8$

Calculation:

Unit vector in the direction of $\text{u}=\langle 3,3\rangle$ is,

  $\begin{array}{l}\text{v}=\left(\frac{1}{\Vert \text{v}\Vert }\right)\text{u}\\ \Rightarrow \text{v}=\frac{1}{\sqrt{3^2+3^2}}\langle 3,3\rangle \\ \Rightarrow \text{v}=\frac{1}{\sqrt{9+9}}\langle 3,3\rangle \\ \Rightarrow \text{v}=\frac{1}{\sqrt{18}}\langle 3,3\rangle \\ \Rightarrow \text{v}=\frac{1}{3\sqrt{2}}\langle 3,3\rangle \\ \Rightarrow \text{v}=\langle \frac{3}{3\sqrt{2}},\frac{3}{3\sqrt{2}}\rangle \\ \Rightarrow \text{v}=\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rangle \end{array}$

Since we need the vector v that has a magnitude of 8 units, so the unit vector v must be multiplied by 8. That is,

  $\begin{array}{l}\text{v}=8\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rangle \\ \Rightarrow \text{v}=\langle \frac{8}{\sqrt{2}},\frac{8}{\sqrt{2}}\rangle \\ \Rightarrow \text{v}=\langle 4\sqrt{2},4\sqrt{2}\rangle \end{array}$