Chapter 9.6, Problem 60AYU

To determine

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

Answer to Problem 60AYU

Solution:

$\begin{array}{l}\text{cos}\alpha = -\frac{3}{7}; \text{cos}\beta = \frac{6}{7}; \text{cos}\gamma = \frac{2}{7}\\ v = 14\left[\left(-\frac{3}{7}\right)i + \left(\frac{6}{7}\right)j + \left(\frac{2}{7}\right)k\right]\end{array}$

Explanation of Solution

Given:

$v = -6i + 12j + 4k$

Calculation:

Direction angles are given by:

$\begin{array}{l}\text{cos}\alpha = \frac{-6}{\sqrt{{(-6)}^2 + {12}^2 + 4^2}} = \frac{-6}{\sqrt{196}} = \frac{-6}{\sqrt{14}} = -\frac{3}{7}\\ \text{cos}\beta = \frac{12}{\sqrt{{(-6)}^2 + {12}^2 + 4^2}} = \frac{12}{\sqrt{196}} = \frac{12}{14} = \frac{6}{7}\\ \text{cos}\gamma = \frac{-2}{\sqrt{{(-6)}^2 + {12}^2 + 4^2}} = \frac{4}{\sqrt{196}} = \frac{4}{14} = \frac{2}{7}\end{array}$

Where we get,

$\left|\right|v\left|\right| = \sqrt{{(-6)}^2 + {12}^2{4}^2} = \sqrt{36 + 144 + 16} = \sqrt{196} = 14$

Then we have,

$\begin{array}{l}v = \left|\right|v\left|\right|[(\text{cos}\alpha )i + (\text{cos}\beta )j + (\text{cos}\gamma )k]\\ v = 14\left[\left(-\frac{3}{7}\right)i + \left(\frac{6}{7}\right)j + \left(\frac{2}{7}\right)k\right]\end{array}$