Direction Angle Of A Vector

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 indicate $Q$ . In symbols the magnitude of $\vec{P Q}$ is written as $| \vec{P Q} |$ .

If the coordinates of the initial bespeak and the finish point of a vector is given, the Distance Formula can exist used to observe its magnitude.

$| \vec{P Q} | = \sqrt{{(x_{2} - 10_{1})}^{2} + {(y_{ii} - y_{i})}^{2}}$

Example i:

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

Solution:

Employ the Altitude Formula.

Substitute the values of $x_{1}$ , $y_{i}$ , $10_{2}$ , and $y_{2}$ .

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

The magnitude of $\vec{P Q}$ is near $4.five$ .

Direction of a Vector

The management of a vector is the measure out of the bending it makes with a horizontal line .

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

$\tan θ = \frac{y}{10}$ , where $x$ is the horizontal change and $y$ is the vertical change

$\tan θ = \frac{y_{2} - y_{i}}{x_{two} - x_{one}}$ , where $(x_{i}, y_{ane})$ is the initial point and $(x_{two}, y_{2})$ is the terminal signal.

Case two:

Discover the management 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 $\tan θ = \frac{y_{two} - y_{1}}{x_{2} - x_{ane}}$ .