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Magnitude and Direction of Vectors

Magnitude of a Vector

The magnitude of a vector $\vec{P Q}$ is the distance between the initial point $P$ and the end point $Q$ . In symbols the magnitude of $\vec{P Q}$ is written as $| \vec{P Q} |$ .

If the coordinates of the initial point and the end point of a vector is given, the Distance Formula can be used to find its magnitude.

Math diagram

$| \vec{P Q} | = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Example 1:

Find the magnitude of the vector $\vec{P Q}$ whose initial point $P$ is at $(1, 1)$ and end point is at $Q$ is at $(5, 3)$ .

Solution:

Use the Distance Formula.

Substitute the values of $x_{1}$ , $y_{1}$ , $x_{2}$ , and $y_{2}$ .

$\begin{array}{l} | \vec{P Q} | = \sqrt{{(5 - 1)}^{2} + {(3 - 1)}^{2}} \\ = \sqrt{4^{2} + 2^{2}} \\ = \sqrt{16 + 4} \\ = \sqrt{20} \\ \approx 4.5 \end{array}$

The magnitude of $\vec{P Q}$ is about $4.5$ .

Direction of a Vector

The direction of a vector is the measure of the angle it makes with a horizontal line .

One of the following formulas can be used to find the direction of a vector:

$\tan θ = \frac{y}{x}$ , where $x$ is the horizontal change and $y$ is the vertical change

$\tan θ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , where $(x_{1}, y_{1})$ is the initial point and $(x_{2}, y_{2})$ is the terminal point.

Example 2:

Find the direction of the vector $\vec{P Q}$ whose initial point $P$ is at $(2, 3)$ and end point is at $Q$ is at $(5, 8)$ .

The coordinates of the initial point and the terminal point are given. Substitute them in the formula $\tan θ = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$\begin{array}{l} \tan θ = \frac{8 - 3}{5 - 2} \\ = \frac{5}{3} \end{array}$

Find the inverse tan, then use a calculator.

$\begin{array}{l} θ = \tan^{- 1} (\frac{5}{3}) \\ \approx 59 ° \end{array}$

The vector $\vec{P Q}$ has a direction of about $59 °$ .