Question

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

$v=j+k$

Short answer

The given vector is

$v=\sqrt{2}(\cos {90}^{\circ }i+\cos {45}^{\circ }j+\cos {45}^{\circ }k)$

Step-by-step solution

Step 1. Given

Vector v as:

$v=j+k$

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,

$$\begin{gathered} v=j+k \\ letsfindthemagnitudeofvi.e. \\ v=\sqrt{0^2+{\left(1\right)}^2+{\left(1\right)}^2} \\ =\sqrt{2} \\ Nowapplyangleformulatogettherespectiveangles \\ \cos \alpha =\frac{0}{\sqrt{2}} \\ \alpha ={\cos }^{-1}\left(\frac{0}{\sqrt{2}}\right) \\ \approx 90 \\ \beta ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ \approx 45 \\ \gamma ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ \approx 45 \\ Thus, \\ v=\sqrt{2}(\cos {90}^{\circ }i+\cos {45}^{\circ }j+\cos {45}^{\circ }k) \end{gathered}$$