Question

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

$v=i+j+k$

Short answer

The given vector is

$v=\sqrt{3}(\cos 54.7^{\circ }i+\cos 54.7^{\circ }j+\cos 54.7^{\circ }k)$

Step-by-step solution

Step 1. Given

Vector v as:

$v=i+j+k$

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,

$$\begin{gathered} v=i+j+k \\ letsfindthemagnitudeofvi.e. \\ v=\sqrt{1^2+{\left(1\right)}^2+{\left(1\right)}^2} \\ =\sqrt{3} \\ Nowapplyangleformulatogettherespectiveangles \\ \cos \alpha =\frac{1}{\sqrt{3}} \\ \alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \approx 54.7 \\ \beta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \approx 54.7 \\ \gamma ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \approx 54.7 \\ Thus, \\ v=\sqrt{3}(\cos 54.7^{\circ }i+\cos 54.7^{\circ }j+\cos 54.7^{\circ }k) \end{gathered}$$