Question

Magnitude and Direction of a Vector

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

Short answer

The magnitude of the vector is$1$and the direction (in degrees) is $\theta ={45}^{°}$

Step-by-step solution

Step 1. Given information

A vector is given as

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-

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

And the direction is-

$\begin{array}{l}\tan \theta =\frac{\frac{-\sqrt{2}}{2}}{\frac{-\sqrt{2}}{2}}\\ \theta ={\tan }^{-1}1\\ ={45}^{°}\end{array}$