Chapter 9, Problem 55RE

To determine

To find: The direction angles of the vector $v=3i-4j+2k$ .

Answer to Problem 55RE

The direction angles of the vector $v=3i-4j+2k$ are $\alpha =56.1°$ , $\beta =138°$ and $\gamma =68.2°$ .

Explanation of Solution

Given information:

The vector is

  $v=3i-4j+2k$

Calculation:

If $v=ai+bj+ck$ is a non zero vector in space, the direction angles $\alpha ,\beta$ and $\gamma$ .

Calculate the magnitude of the vector.

  $\begin{array}{c}\Vert v\Vert =\sqrt{a^2+b^2+c^2}\\ =\sqrt{3^2+{\left(-4\right)}^2+2^2}\\ =\sqrt{9+16+4}\\ =\sqrt{29}\end{array}$

Calculate the direction angle $\alpha$ .

  $\begin{array}{c}\cos \alpha =\frac{a}{\Vert v\Vert }\\ \alpha ={\cos }^{-1}\left(\frac{3}{\sqrt{29}}\right)\\ \alpha =56.1°\end{array}$

Calculate the direction angle $\beta$ .

  $\begin{array}{c}\cos \beta =\frac{b}{\Vert v\Vert }\\ \beta ={\cos }^{-1}\left(\frac{-4}{\sqrt{29}}\right)\\ \beta =138°\end{array}$

Calculate the direction angle $\gamma$ .

  $\begin{array}{c}\cos \gamma =\frac{b}{\Vert v\Vert }\\ \gamma ={\cos }^{-1}\left(\frac{2}{\sqrt{29}}\right)\\ \gamma =68.2°\end{array}$

Therefore, the direction angles of the vector $v=3i-4j+2k$ are $\alpha =56.1°$ , $\beta =138°$ and $\gamma =68.2°$ .