Question

Magnitude and Direction of a Vector

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

$\mathbf{v}=⟨40,9⟩$

Short answer

The magnitude of the vector is $41$and the direction (in degrees) is$\theta ={12.68}^{°}$

Step-by-step solution

Step 1. Given information

A vector is given as$\mathbf{v}=⟨40,9⟩$

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-

$\begin{array}{l}\left|v\right|=\sqrt{{40}^2+9^2}\\ =\sqrt{1600+81}\\ =\sqrt{1681}\\ =41\end{array}$

And the direction is-

$\begin{array}{l}\tan \theta =\frac{9}{40}\\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{9}{40}\right)\\ \Rightarrow \theta ={12.68}^{°}\end{array}$