# 22.1: Introduction to Vectors

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There is yet another way to discuss the $$2$$-dimensional plane, which highlights some other useful properties of the plane $$\mathbb{R}^2$$. This is the notion of $$\mathbb{R}^2$$ as a vector space. We start by defining vectors in the plane.

## Definition: Geometric Vector

A geometric vector in the plane is a geometric object in the plane $$\mathbb{R}^2$$ that is given by a direction and magnitude. We denote a vector by $$\vec{v}$$ (it is written by some authors as v), its magnitude is denoted by $$||\vec{v}||$$, and its directional angle by $$\theta$$.

Vectors are often represented by directed line segments $$\vec{v}=\overrightarrow{PQ}$$, where two line segments represent the same vector, if one can be moved to the other by parallel translation (without changing its direction or magnitute).

In particular, we can always represent a vector $$\vec{v}$$ by $$\overrightarrow{OR}$$ by arranging the starting point of $$\vec{v}$$ to the origin $$O(0,0)$$. If $$R$$ is given in coordinates by $$R(a,b)$$, then we also write for $$\vec{v}=\overrightarrow{OR}$$,

$\boxed{\vec{v}=\langle a,b\rangle} \quad\text{ or, alternatively, }\quad \boxed{\vec{v}=\begin{bmatrix} a \\ b \end{bmatrix}} \nonumber$

## Example $$\PageIndex{1}$$

Graph the vectors $$\vec{v}, \vec{w}, \vec{r}, \vec{s}, \vec{t}$$ in the plane, where $$\vec{v}=\overrightarrow{PQ}$$ with $$P(6,3)$$ and $$Q(4,-2)$$, and

$\vec{w}=\langle 3, -1\rangle, \quad \vec{r}=\langle -4, -2\rangle, \quad \vec{s}=\langle 0, 2\rangle, \quad \vec{t}=\langle -5, 3\rangle \nonumber$

Solution

The formulas for the magnitude and the directional angle of a vector can be obtained precisely the same way as the absolute value and angle of a complex number. From equation 21.1.3 in Observation [OBS:polar-form], we therefore obtain the following analogue formulas.

## Observation: Magnitude and angle of vector

Let $$\vec{v}=\langle a,b\rangle$$ be a vector in the plane $$\mathbb{R}^2$$. Then the magnitude and angle of $$\vec{v}$$ are given by:

$\label{EQU:magnitude-and-angle} \boxed{ ||\vec{v}||=\sqrt{a^2+b^2} } \quad \text{ and } \quad \boxed{\tan(\theta)=\dfrac{b}{a}}$

Conversely, we can recover the coordinates of a vector $$\vec{v}$$ from its magnitude $$||\vec{v}||$$ and angle $$\theta$$ by (see equation 21.1.1 from page ):

$\label{EQU:magnitude-angle-to-component} \boxed{ \vec{v}=\langle \,\,\, ||\vec{v}||\cdot \cos(\theta)\,\,\, , \,\,\, ||\vec{v}||\cdot \sin(\theta) \,\,\, \rangle}$

## Example $$\PageIndex{2}$$

Find the magnitude and directional angle of the given vectors.

1. $$\langle -6,6\rangle$$
2. $$\langle 4,-3\rangle$$
3. $$\langle -2\sqrt{3},-2\rangle$$
4. $$\langle 8, 4\sqrt{5} \rangle$$
5. $$\overrightarrow{PQ}$$, where $$P(9,2)$$ and $$Q(3,10)$$

Solution

1. We use formulas $$\ref{EQU:magnitude-and-angle}$$, and the calculation is in analogy with Example 21.1.3. The magnitude of $$\vec{v}=\langle -6,6\rangle$$ is

$||\vec{v}||=\sqrt{(-6)^2+6^2}=\sqrt{36+36}=\sqrt{72}=\sqrt{36\cdot 2}=6\sqrt{2} \nonumber$

The directional angle $$\theta$$ is given by $$\tan(\theta)=\dfrac{6}{-6}=-1$$. Now, since $$\tan^{-1}(-1)=-\tan^{-1}(1)=-45^\circ$$ is in the fourth quadrant, but $$\vec{v}=\langle -6,6\rangle$$ drawn at the origin $$O(0,0)$$ has its endpoint in the second quadrant, we see that the angle $$\theta=-45^\circ+180^\circ=135^\circ$$.

1. The magnitude of $$\vec{v}=\langle4,-3\rangle$$ is

$||\vec{v}||=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5 \nonumber$

The directional angle is given by $$\tan(\theta)=\dfrac{-3}{4}$$. Since $$\tan^{-1}\left(\dfrac{-3}{4}\right)\approx -36.9^\circ$$ is in the fourth quadrant, and $$\vec{v}=\langle4,-3\rangle$$ is in the fourth quadrant, we see that

$\theta=\tan^{-1}\left(\dfrac{-3}{4}\right)\approx -36.9^\circ \nonumber$

1. The magnitude of $$\vec{v}=\langle -2\sqrt{3},-2\rangle$$ is

$||\vec{v}||=\sqrt{(-2\sqrt{3})^2+(-2)^2}=\sqrt{4\cdot 3+4}=\sqrt{12+4}=\sqrt{16}=4 \nonumber$

The directional angle is given by $$\tan(\theta)=\dfrac{-2}{-2\sqrt{3}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$$. Now, $$\tan^{-1}\left(\dfrac{\sqrt{3}}{3}\right)=30^\circ=\dfrac{\pi}{6}$$ is in the first quadrant, whereas $$\vec{v}=\langle -2\sqrt{3},-2\rangle$$ is in the third quadrant. Therefore, the angle is given by adding an additional $$\pi$$ to the angle.

$\theta=\pi+\dfrac{\pi}{6}+\dfrac{6\pi+\pi}{6}=\dfrac{7\pi}{6} \nonumber$

1. The magnitude of $$\vec{v}=\langle 8, 4\sqrt{5} \rangle$$ is

$||\vec{v}||=\sqrt{8^2+(4\sqrt{5})^2}=\sqrt{64+16\cdot 5}=\sqrt{64+80}=\sqrt{144}=12 \nonumber$

The directional angle is given by the equation $$\tan(\theta)=\dfrac{4\sqrt{5}}{8}=\dfrac{\sqrt{5}}{2}$$. Since both $$\tan^{-1}\left(\dfrac{\sqrt{5}}{2}\right)\approx 48.2^\circ$$ and the endpoint of $$\vec{v}$$ (represented with beginning point at the origin) are in the first quadrant, we have:

$\theta=\tan^{-1}\left(\dfrac{\sqrt{5}}{2}\right)\approx 48.2^\circ \nonumber$

1. We first need to find the vector $$\vec{v}=\overrightarrow{PQ}$$ in the form $$\vec{v}=\langle a,b \rangle$$. The vector in the plane below shows that $$\vec{v}$$ is given by $\vec{v}=\langle 3-9, 10-2 \rangle =\langle -6,8 \rangle \nonumber$

From this we calculate the magnitude to be

$||\vec{v}|| = \sqrt{(-6)^2+8^2}=\sqrt{36+64}=\sqrt{100}=10 \nonumber$

The directional angle is given by $$\tan(\theta)=\dfrac{8}{-6}=-\dfrac{4}{3}$$. Now, $$\tan^{-1}\left(-\dfrac{4}{3}\right)\approx -53.1^\circ$$ is in quadrant IV, whereas $$\vec{v}=\langle-6,8\rangle$$ has an endpoint in quadrant II (when representing $$\vec{v}$$ with starting point at the origin $$O(0,0)$$). Therefore, the directional angle is

$\theta=180^\circ+\tan^{-1}\left(-\dfrac{4}{3}\right)\approx 126.9^\circ \nonumber$

This page titled 22.1: Introduction to Vectors is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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