Chapter 9.6, Problem 63AYU

In Problems 59-66, find the direction angles of each vector. Write each vector in the form of equation (7).

$v=i+j$

To determine

To find: The direction angles of each vector. Write the vector in terms of its magnitude and direction cosines.

Answer to Problem 63AYU

Solution:

$\begin{array}{l}\text{cos}\alpha = \frac{1}{\sqrt{2}}; \text{cos}\beta = \frac{1}{\sqrt{2}}; \text{cos}\gamma = 0\\ v = \sqrt{2}\left[\left(\frac{1}{\sqrt{2}}\right)i + \left(\frac{1}{\sqrt{2}}\right)j\right]\end{array}$

Explanation of Solution

Given:

$v = i + j$

Calculation:

Direction angles are given by:

$\begin{array}{l}\text{cos}\alpha = \frac{1}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{1}{\sqrt{2}}\\ \text{cos}\beta = \frac{1}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{1}{\sqrt{2}}\\ \text{cos}\gamma = \frac{0}{\sqrt{1^2 + 1^2 + 0^2}} = 0\end{array}$

Where we get, $\left|\right|v\left|\right| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2}$

Then we have,

$\begin{array}{l}v = \left|\right|v\left|\right|[(\text{cos}\alpha )i + (\text{cos}\beta )j + (\text{cos}\gamma )k]\\ v = \sqrt{2}\left[\left(\frac{1}{\sqrt{2}}\right)i + \left(\frac{1}{\sqrt{2}}\right)j + 0k\right]\\ v = \sqrt{2}\left[\left(\frac{1}{\sqrt{2}}\right)i + \left(\frac{1}{\sqrt{2}}\right)j\right]\end{array}$