Chapter 6.3, Problem 79E

To determine

To Find:The magnitude vector and direction angle of the vector v

  $v=12i+15j$

Answer to Problem 79E

The magnitude of the vector is $3\sqrt{41}$ and the direction angle is ${51.34}^{\circ }$

Explanation of Solution

Given:Find the magnitude vector and direction angle of the vector v

  $v=12i+15j$

FORMULA USED:

Magnitude of the vector : $\Vert \overrightarrow{v}\Vert =\sqrt{v_1{}^2+v_2{}^2}$

Given $v=12i+15j$

The magnitude of the vector is given by $\Vert \overrightarrow{v}\Vert =\sqrt{v_1{}^2+v_2{}^2}$

Thus

  $\Vert \overrightarrow{v}\Vert =\sqrt{{12}^2+{15}^2}=\sqrt{369}=3\sqrt{41}$

Now given

  $v=ai+bj$

The directive angle is given by

  $\begin{array}{l}\left|\arctan \frac{b}{a}\right|\\ i.e\\ \theta =\left|\arctan \frac{15}{12}\right|\approx \left|{51.34}^{\circ }\right|={51.34}^{\circ }\end{array}$

Hence, The magnitude of the vector is $3\sqrt{41}$ and the direction angle is ${51.34}^{\circ }$

  $\begin{array}{l}v=12i+15j\\ \end{array}$