Question

Find the direction angles of the vector$v=3i-4j+2k.$

Short answer

The direction angles of the vector are$\alpha \approx 56.1°,\beta \approx 138°,and\gamma \approx 68.2°.$

Step-by-step solution

Step 1. Given Information

The given vector is$v=3i-4j+2k.$

Step 2. Finding the direction angles

To find the direction angles $\alpha ,\beta ,and\gamma$use the formula

$$\begin{gathered} \cos \alpha =\frac{a}{\left|\left|v\right|\right|},\cos \beta =\frac{b}{\left|\left|v\right|\right|},\cos \gamma =\frac{c}{\left|\left|v\right|\right|} \\ where\left|\left|v\right|\right|=\sqrt{a^2+b^2+c^2} \end{gathered}$$

Let's first find $\left|\left|v\right|\right|$

So,$$\begin{gathered} \left|\left|v\right|\right|=\sqrt{3^2+{\left(-4\right)}^2+2^2} \\ \left|\left|v\right|\right|=\sqrt{29} \end{gathered}$$

Step 3. Solve

Proceed with the calculation further by using the formula.

Thus,

$$\begin{gathered} \cos \alpha =\frac{3}{\sqrt{29}} \\ \mathrm{Rationalize}\mathrm{the}\mathrm{denominator} \\ \mathrm{cos\alpha }=\frac{3\sqrt{29}}{29} \\ \mathrm{\alpha }={\cos }^{-1}\left(\frac{3\sqrt{29}}{29}\right) \\ \mathrm{\alpha }\approx 56.1° \end{gathered}$$

And

$$\begin{gathered} \cos \beta =\frac{-4}{\sqrt{29}} \\ \mathrm{Rationalize}\mathrm{the}\mathrm{denominator} \\ \mathrm{cos\beta }=-\frac{4\sqrt{29}}{29} \\ \mathrm{\beta }={\cos }^{-1}\left(-\frac{4\sqrt{29}}{29}\right) \\ \mathrm{\beta }\approx 138° \end{gathered}$$

And

$$\begin{gathered} \cos \gamma =\frac{2}{\sqrt{29}} \\ \mathrm{Rationalize}\mathrm{the}\mathrm{denominator} \\ \mathrm{cos\gamma }=\frac{2\sqrt{29}}{29} \\ \mathrm{\gamma }={\cos }^{-1}\left(\frac{2\sqrt{29}}{29}\right) \\ \mathrm{\gamma }\approx 68.2° \end{gathered}$$