Chapter 9.6, Problem 66AYU

To determine

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

Answer to Problem 66AYU

Solution:

$\begin{array}{l}\text{cos}\alpha = \frac{2}{\sqrt{29}}; \text{cos}\beta = \frac{3}{\sqrt{29}}; \text{cos}\gamma = -\frac{4}{\sqrt{29}}\\ v = \sqrt{29}\left[\left(\frac{2}{\sqrt{29}}\right)i + \left(\frac{3}{\sqrt{29}}\right)j + \left(-\frac{4}{\sqrt{29}}\right)k\right]\end{array}$

Explanation of Solution

Given:

$v = 2i + 3j - 4k$

Calculation:

Direction angles are given by:

$\begin{array}{l}\text{cos}\alpha = \frac{2}{\sqrt{2^2 + 3^2 + {(-4)}^2}} = \frac{2}{\sqrt{4 + 9 + 16}} = \frac{2}{\sqrt{29}}\\ \text{cos}\beta = \frac{3}{\sqrt{2^2 + 3^2 + {(-4)}^2}} = \frac{3}{\sqrt{4 + 9 + 16}} = \frac{3}{\sqrt{29}}\\ \text{cos}\gamma = \frac{-4}{\sqrt{2^2 + 3^2 + {(-4)}^2}} = \frac{-4}{\sqrt{4 + 9 + 16}} = \frac{-4}{\sqrt{29}}\end{array}$

Where we get, $\left|\right|v\left|\right| = \sqrt{2^2 + 3^2 + {(-4)}^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$

Then we have,

$\begin{array}{l}v = \left|\right|v\left|\right|[(\text{cos}\alpha )i + (\text{cos}\beta )j + (\text{cos}\gamma )k]\\ v = \sqrt{29}\left[\left(\frac{2}{\sqrt{29}}\right)i + \left(\frac{3}{\sqrt{29}}\right)j + \left(-\frac{4}{\sqrt{29}}\right)k\right]\end{array}$