Chapter 9.6, Problem 61AYU

To determine

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

Answer to Problem 61AYU

Solution:

$\begin{array}{l}\text{cos}\alpha = \frac{1}{\sqrt{3}}; \text{cos}\beta = \frac{1}{\sqrt{3}}; \text{cos}\gamma = \frac{1}{\sqrt{3}}\\ v = \sqrt{3}\left[\left(\frac{1}{\sqrt{3}}\right)i + \left(\frac{1}{\sqrt{3}}\right)j + \left(\frac{1}{\sqrt{3}}\right)k\right]\end{array}$

Explanation of Solution

Given:

$v = i + j + k$

Calculation:

Direction angles are given by:

$\begin{array}{l}\text{cos}\alpha = \frac{1}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}}\\ \text{cos}\beta = \frac{1}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}}\\ \text{cos}\gamma = \frac{1}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}}\end{array}$

Where we get, $\left|\right|v\left|\right| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

Then we have,

$\begin{array}{l}v = \left|\right|v\left|\right|[(\text{cos}\alpha )i + (\text{cos}\beta )j + (\text{cos}\gamma )k]\\ v = \sqrt{3}\left[\left(\frac{1}{\sqrt{3}}\right)i + \left(\frac{1}{\sqrt{3}}\right)j + \left(\frac{1}{\sqrt{3}}\right)k\right]\end{array}$