Question

Magnitude and Direction of a Vector

Find the magnitude and direction (in degrees) of the vector.

$\mathbf{v}=\mathbf{i}+\sqrt{3}\mathbf{j}$

Short answer

The magnitude of the vector is$2$and the direction (in degrees) is $\theta ={60}^{°}$

Step-by-step solution

Step 1. Given information

A vector is given as $\mathbf{v}=\mathbf{i}+\sqrt{3}\mathbf{j}$

Step 2. Concept used

The formula used to calculate the length or magnitude of a vector is-

$\left|v\right|=\sqrt{a_1^2+a_2^2}$

And for the direction or angle, we use-

$\tan \theta =\frac{a_2}{a_1}$

Use the formulas to get the required answers.

Step 3. Calculation

The magnitude of vector is-

$\begin{array}{l}\left|v\right|=\sqrt{1^2+{\left(\sqrt{3}\right)}^2}\\ =\sqrt{1+3}\\ =\sqrt{4}\\ =2\end{array}$

And the direction is-

$\begin{array}{l}\tan \theta =\frac{\sqrt{3}}{1}\\ \theta ={\tan }^{-1}\sqrt{3}\\ ={60}^{°}\end{array}$