Question

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

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

Short answer

The given vector is

$v=7(\cos 115.4^{\circ }i+\cos 31.0^{\circ }j+\cos 73.4^{\circ }k)$

Step-by-step solution

Step 1. Given

Vector v as:

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

Step 2. Concept

For a given vector in the xyz plane.

A directional angle is defined as :

$$\begin{gathered} \cos \alpha =\frac{a}{\left|v\right|}\angle \alpha anglebetweenvandxaxis. \\ \cos \beta =\frac{b}{\left|v\right|}\angle \beta anglebetweenvandyaxis. \\ \cos \gamma =\frac{c}{\left|v\right|}\angle \gamma anglebetweenvandzaxis. \end{gathered}$$

Step 3. Calculation

Consider,

$$\begin{gathered} v=-6i+12j+4k \\ letsfindthemagnitudeofvi.e. \\ v=\sqrt{{(-6)}^2+{\left(12\right)}^2+{\left(4\right)}^2} \\ =\sqrt{196} \\ =14 \\ Nowapplyangleformulatogettherespectiveangles \\ \cos \alpha =\frac{-6}{14} \\ \alpha ={\cos }^{-1}\left(-\frac{3}{7}\right) \\ \approx 115.4 \\ \beta ={\cos }^{-1}\left(\frac{12}{14}\right) \\ \approx 31.0 \\ \gamma ={\cos }^{-1}\left(\frac{4}{14}\right) \\ \approx 73.4 \\ Thus, \\ v=7(\cos 115.4^{\circ }i+\cos 31.0^{\circ }j+\cos 73.4^{\circ }k) \end{gathered}$$