Chapter 9, Problem 106RE

To determine

To calculate: The direction angles of the vector $v=i\text{}-j+2k$ .

Answer to Problem 106RE

The direction angles of the vector $v=i\text{}-j+2k$ are $\alpha =65.905158^{°},\beta =114.094843^{°}and\gamma =35.265^{°}$

Explanation of Solution

Given information:

The vectors are given as:

  $v=i\text{}-j+2k$

Formula used:

let $v=ai+bj+ck$ is vector,

Then, direction angles $\text{α}\text{,}\text{β}\text{and}\text{γ}$ are given as,

  $cos\alpha =\frac{a}{\Vert v\Vert },cos\beta =\frac{b}{\Vert v\Vert }andcos\gamma =\frac{c}{\Vert v\Vert }$

Where $\Vert v\Vert =\sqrt{a^2+b^2+c^2}$

Calculation:

Consider the vectors,

  $v=i\text{}-j+2k$

The value of $a=1,b=-1\text{and}\text{}c\text{}=\text{}2$

Further,

  $\begin{array}{l}cos\alpha =\frac{a}{\sqrt{a^2+b^2+c^2}}\\ cos\alpha =\frac{1}{\sqrt{1^2+-1^2+2^2}}\\ cos\alpha =\frac{1}{\sqrt{6}}\end{array}$

The value of $\alpha$ is $65.905158^{°}$

Furthermore,

  $\begin{array}{l}cos\beta =\frac{b}{\sqrt{a^2+b^2+c^2}}\\ cos\beta =\frac{-1}{\sqrt{1^2+-1^2+2^2}}\\ cos\beta =\frac{-1}{\sqrt{6}}\end{array}$

The value of $\beta \text{}$ is $114.094843^{°}$

And

  $\begin{array}{l}cos\gamma =\frac{c}{\sqrt{a^2+b^2+c^2}}\\ cos\gamma =\frac{2}{\sqrt{1^2+-1^2+2^2}}\\ cos\gamma =\frac{2}{\sqrt{6}}\end{array}$

The value of $\gamma$ is $35.265^{°}$

Thus, the direction angles of the vector $v=i\text{}-j+2k$ are $\alpha =65.905158^{°},\beta =114.094843^{°}and\gamma =35.265^{°}$