# 22.2: Operations on vectors

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There are two basic operations on vectors, which are the scalar multiplication and the vector addition. We start with the scalar multiplication.

## Definition: Scalar Multiplication

The scalar multiplication of a real number $$r$$ with a vector $$\vec{v}=\langle a,b\rangle$$ is defined to be the vector given by multiplying $$r$$ to each coordinate.

$\boxed{ r\cdot \langle a,b \rangle := \langle r\cdot a, r \cdot b \rangle }$

We study the effect of scalar multiplication with an example.

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

Multiply, and graph the vectors.

1. $$4\cdot \langle-2,1\rangle$$
2. $$(-3)\cdot \langle -6,-2 \rangle$$

Solution

1. The calculation is straightforward.

$4\cdot \langle-2,1\rangle=\langle4\cdot (-2),4\cdot 1\rangle=\langle-8,4\rangle \nonumber$

The vectors are displayed below. We see, that $$\langle-2,1\rangle$$ and $$\langle-8,4\rangle$$ both have the same directional angle, and the latter stretches the former by a factor $$4$$.

1. Algebraically, we calculate the scalar multiplication as follows:

$(-3)\cdot \langle -6,-2 \rangle= \langle (-3)\cdot(-6),(-3)\cdot(-2) \rangle =\langle 18,6 \rangle \nonumber$

Furthermore, graphing the vectors gives:

We see that the directional angle of the two vectors differs by $$180^\circ$$. Indeed, $$\langle18,6\rangle$$ is obtained from $$\langle -6,-2\rangle$$ by reflecting it at the origin $$O(0,0)$$ and then stretching the result by a factor $$3$$.

We see from the above example, that scalar multiplication by a positive number $$c$$ does not change the angle of the vector, but it multiplies the magnitude by $$c$$.

## Observation: c-times-vector

Let $$\vec{v}$$ be a vector with magnitude $$||\vec{v}||$$ and angle $$\theta_{\vec{v}}$$. Then, for a positive scalar, $$r>0$$, the scalar multiple $$r\cdot \vec{v}$$ has the same angle as $$\vec{v}$$, and a magnitude that is $$r$$ times the magnitude of $$\vec{v}$$:

$\text{for r>0: } \quad ||r\cdot \vec{v}|| = r\cdot ||\vec{v}|| \quad\text{ and } \quad \theta_{r\cdot \vec{v}}=\theta_{\vec{v}}$

## Definition: Unit Vector

A vector $$\vec{u}$$ is called a unit vector, if it has a magnitude of $$1$$.

$\vec{u} \text{ is a unit vector } \quad \iff \quad ||\vec{u}||=1$

There are two special unit vectors, which are the vectors pointing in the $$x$$- and the $$y$$-direction.

$\boxed{\vec{i}:=\langle 1,0\rangle}\quad \text{ and }\quad \boxed{\vec{j}:=\langle 0,1\rangle}$

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

Find a unit vector in the direction of $$\vec{v}$$.

1. $$\langle 8, 6 \rangle$$
2. $$\langle -2,3\sqrt{7} \rangle$$

Solution

1. Note that the magnitude of $$\vec{v}=\langle 8, 6 \rangle$$ is

$||\langle 8, 6 \rangle||=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10 \nonumber$

Therefore, if we multiply $$\langle 8, 6 \rangle$$ by $$\dfrac{1}{10}$$ we obtain $$\dfrac{1}{10}\cdot \langle 8, 6 \rangle=\langle \dfrac{8}{10}, \dfrac{6}{10} \rangle=\langle \dfrac{4}{5}, \dfrac{3}{5} \rangle$$, which (according to Observation [OBS:c-times-vector] above) has the same directional angle as $$\langle 8, 6 \rangle$$, and has a magnitude of $$1$$:

$\left|\left|\dfrac{1}{10}\cdot \langle 8, 6 \rangle\right|\right| = \dfrac{1}{10}\cdot ||\langle 8, 6 \rangle||=\dfrac{1}{10}\cdot 10=1 \nonumber$

1. The magnitude of $$\langle -2,3\sqrt{7} \rangle$$ is

$||\langle -2,3\sqrt{7} \rangle||=\sqrt{(-2)^2+(3\sqrt{7})^2}=\sqrt{4+9\cdot 7}=\sqrt{4+63}=\sqrt{67} \nonumber$

Therefore, we have the unit vector

$\dfrac{1}{\sqrt{67}}\cdot \langle -2,3\sqrt{7} \rangle=\langle \dfrac{-2}{\sqrt{67}},\frac{3\sqrt{7} }{\sqrt{67}}\rangle=\langle \dfrac{-2\sqrt{67}}{67},\dfrac{3\sqrt{7}\sqrt{67}}{67}\rangle=\langle \dfrac{-2\sqrt{67}}{67},\dfrac{3\sqrt{469}}{67}\rangle \nonumber$

which again has the same directional angle as $$\langle -2,3\sqrt{7} \rangle$$. Now, $$\dfrac{1}{\sqrt{67}}\cdot \langle -2,3\sqrt{7} \rangle$$ is a unit vector, since $||\dfrac{1}{\sqrt{67}}\cdot \langle -2,3\sqrt{7} \rangle||=\dfrac{1}{\sqrt{67}}\cdot ||\langle -2,3\sqrt{7} \rangle||=\dfrac{1}{\sqrt{67}}\cdot \sqrt{67}=1 \nonumber$

The second operation on vectors is vector addition, which we discuss now.

Let $$\vec{v}=\langle a,b \rangle$$ and $$\vec{w}=\langle c,d \rangle$$ be two vectors. Then the vector addition $$\vec{v}+\vec{w}$$ is defined by component-wise addition:

$\boxed{\,\, \langle a,b \rangle+\langle c,d \rangle:=\langle a+c,b+d \rangle \,\,}$

In terms of the plane, the vector addition corresponds to placing the vectors $$\vec{v}$$ and $$\vec{w}$$ as the edges of a parallelogram, so that $$\vec{v}+\vec{w}$$ becomes its diagonal. This is displayed below.

## Example $$\PageIndex{3}$$

Perform the vector addition and simplify as much as possible.

1. $$\langle 3,-5 \rangle+\langle 6,4 \rangle$$
2. $$5\cdot \langle -6,2 \rangle-7\cdot \langle 1, -3 \rangle$$
3. $$4\vec{i}+9\vec{j}$$
4. Find $$2\vec{v}+3\vec{w}$$ for $$\vec{v}=-6\vec{i}-4\vec{j}$$ and $$\vec{w}=10\vec{i}-7\vec{j}$$
5. Find $$-3\vec{v}+5\vec{w}$$ for $$\vec{v}=\langle 8,\sqrt{3} \rangle$$ and $$\vec{w}=\langle 0,4\sqrt{3} \rangle$$

Solution

We can find the answer by direct algebraic computation.

1. $$\langle 3,-5\rangle+\langle 6,4\rangle=\langle 3+6,-5+4\rangle=\langle 9,-1\rangle$$
2. $$5 \cdot\langle-6,2\rangle-7 \cdot\langle 1,-3\rangle=\langle-30,10\rangle+\langle-7,21\rangle=\langle-37,31\rangle$$
3. $$4 \vec{i}+9 \vec{j}=4 \cdot\langle 1,0\rangle+9 \cdot\langle 0,1\rangle=\langle 4,0\rangle+\langle 0,9\rangle=\langle 4,9\rangle$$

From the last calculation, we see that every vector can be written as a linear combination of the vectors $$\vec{i}$$ and $$\vec{j}$$.

$\boxed{\langle a,b \rangle = a\cdot \vec{i}+b\cdot \vec{j}} \nonumber$

We will use this equation for the next example (d). Here, $$\vec{v}=-6\vec{i}-4\vec{j}=\langle-6,-4\rangle$$ and $$\vec{w}=10\vec{i}-7\vec{j}=\langle 10,-7\rangle$$. Therefore, we obtain:

1. $$2 \vec{v}+3 \vec{w}=2 \cdot\langle-6,-4\rangle+3 \cdot\langle 10,-7\rangle=\langle-12,-8\rangle+\langle 30,-21\rangle=\langle 18,-29\rangle$$
1. \begin{aligned}-3 \vec{v}+5 \vec{w} &=-3 \cdot\langle 8, \sqrt{3}\rangle+5 \cdot\langle 0,4 \sqrt{3}\rangle\\&=\langle-24,-3 \sqrt{3}\rangle+\langle 0,20 \sqrt{3}\rangle \\&=\langle-24,17 \sqrt{3}\rangle\end{aligned}

Note that the answer could also be written as $$-3\vec{v}+5\vec{w}=-24\vec{i}+17\sqrt{3}\vec{j}$$.

In many applications in the sciences, vectors play an important role, since many quantities are naturally described by vectors. Examples for this in physics include the velocity $$\vec{v}$$, acceleration $$\vec{a}$$, and the force $$\vec{F}$$ applied to an object. [EXA:addition-of-forces] The forces $$\vec{F_1}$$ and $$\vec{F_2}$$ are applied to an object.

## Example $$\PageIndex{4}$$

Find the resulting total force $$\vec{F}=\vec{F_1}+\vec{F_2}$$. Determine the magnitude and directional angle of the total force $$\vec{F}$$. Approximate these values as necessary. Recall that the international system of units for force is the newton $$[1N=1\frac{kg\cdot m}{s^2}]$$.

1. $$\vec{F_1}$$ has magnitude $$3$$ newtons, and angle $$\theta_1=45^\circ$$, $$\vec{F_2}$$ has magnitude $$5$$ newtons, and angle $$\theta_2=135^\circ$$
2. $$||\vec{F_1}||=7$$ newtons, and $$\theta_1=\dfrac{\pi}{6}$$, and $$||\vec{F_2}||=4$$ newtons, and $$\theta_2=\dfrac{5\pi}{3}$$

Solution

1. The vectors $$\vec{F_1}$$ and $$\vec{F_2}$$ are given by their magnitudes and directional angles. However, the addition of vectors (in Definition [DEF:vector-addition]) is defined in terms of their components. Therefore, our first task is to find the vectors in component form. As was stated in equation 22.1.2 [EQU:magnitude-angle-to-component] on page , the vectors are calculated by $$\vec{v}=\langle \,\,\, ||\vec{v}||\cdot \cos(\theta)\,\,\, , \,\,\, ||\vec{v}||\cdot \sin(\theta) \,\,\, \rangle$$. Therefore,

\begin{aligned} \vec{F_1} &= \langle 3\cdot \cos(45^\circ), 3\cdot \sin(45^\circ) \rangle= \langle 3\cdot \dfrac{\sqrt{2}}{2}, 3\cdot \dfrac{\sqrt{2}}{2} \rangle= \langle \dfrac{3\sqrt{2}}{2}, \dfrac{3\sqrt{2}}{2} \rangle \\ \vec{F_2} &= \langle 5\cdot \cos(135^\circ), 5\cdot \sin(135^\circ) \rangle= \langle 5\cdot \dfrac{-\sqrt{2}}{2}, 5\cdot \dfrac{\sqrt{2}}{2} \rangle =\langle \dfrac{-5\sqrt{2}}{2}, \dfrac{5\sqrt{2}}{2} \rangle\end{aligned} \nonumber

The total force is the sum of the forces.

\begin{aligned} \vec{F}&=\vec{F_1}+\vec{F_2}\\& = \langle \dfrac{3\sqrt{2}}{2}, \dfrac{3\sqrt{2}}{2} \rangle+\langle \dfrac{-5\sqrt{2}}{2}, \dfrac{5\sqrt{2}}{2} \rangle \\ &= \langle \dfrac{3\sqrt{2}}{2}+\dfrac{-5\sqrt{2}}{2}, \dfrac{3\sqrt{2}}{2} +\dfrac{5\sqrt{2}}{2}\rangle\\& =\langle \dfrac{3\sqrt{2}-5\sqrt{2}}{2}, \dfrac{3\sqrt{2}+5\sqrt{2}}{2}\rangle \\ &= \langle \dfrac{-2\sqrt{2}}{2}, \dfrac{8\sqrt{2}}{2}\rangle\\& = \langle -\sqrt{2}, 4\sqrt{2}\rangle \end{aligned} \nonumber

The total force applied in components is $$\vec{F}=\langle -\sqrt{2}, 4\sqrt{2}\rangle$$. It has a magnitude of $$||\vec{F}||=\sqrt{(-\sqrt{2})^2+(4\sqrt{2})^2}=\sqrt{2+16\cdot 2}=\sqrt{34}\approx 5.83$$ newton. The directional angle is given by $$\tan(\theta)=\dfrac{4\sqrt{2}}{-\sqrt{2}}=-4$$. Since $$\tan^{-1}(-4)\approx -76.0^\circ$$ is in quadrant IV, and $$\vec{F}=\langle -\sqrt{2}, 4\sqrt{2} \rangle$$ is in quadrant II, we see that the directional angle of $$\vec{F}$$ is

$\theta=180^\circ+\tan^{-1}(-4)\approx180^\circ-76.0^\circ \approx 104^\circ \nonumber$

1. We solve this example in much the same way we solved part (a). First, $$\vec{F_1}$$ and $$\vec{F_2}$$ in components is given by

\begin{aligned} \vec{F_1} &= \langle 7\cdot \cos\Big(\dfrac{\pi}{6}\Big), 7\cdot \sin\Big(\dfrac{\pi}{6}\Big) \rangle= \langle 7\cdot \dfrac{\sqrt{3}}{2}, 7\cdot \dfrac{1}{2} \rangle= \langle \dfrac{7\sqrt{3}}{2}, \dfrac{7}{2} \rangle \\ \vec{F_2} &=\langle 4\cdot \cos\Big(\dfrac{5\pi}{3}\Big), 4\cdot \sin\Big(\dfrac{5\pi}{3}\Big) \rangle= \langle 4\cdot \dfrac{1}{2}, 4\cdot \dfrac{-\sqrt{3}}{2} \rangle= \langle 2, -2\sqrt{3} \rangle\end{aligned} \nonumber

The total force is therefore:

\begin{aligned} \vec{F}&= \vec{F_1}+\vec{F_2}\\&=\langle \dfrac{7\sqrt{3}}{2}, \dfrac{7}{2} \rangle +\langle 2, -2\sqrt{3} \rangle\\&=\langle \dfrac{7\sqrt{3}}{2}+2, \dfrac{7}{2}-2\sqrt{3} \rangle \\ &\approx \langle 8.06, 0.04 \rangle\end{aligned} \nonumber

The magnitude is approximately

$||\vec{F}||\approx \sqrt{(8.06)^2+(0.04)^2}\approx 8.06 \text{ newton} \nonumber$

The directional angle is given by $$\tan(\theta)\approx \dfrac{0.04}{8.06}$$. Since $$\vec{F}$$ is in quadrant I, we see that $$\theta\approx \tan^{-1}\left(\dfrac{0.04}{8.06}\right)\approx 0.3^\circ$$.

## Note

There is an abstract notion of a vector space. Although we do not use this structure for any further computations, we will state its definition. A vector space is a set $$V$$, with the following structures and properties. The elements of $$V$$ are called vectors, denoted by the usual symbol $$\vec{v}$$. For any vectors $$\vec{v}$$ and $$\vec{w}$$ there is a vector $$\vec{v}+\vec{w}$$, called the vector addition. For any real number $$r$$ and vector $$\vec{v}$$, there is a vector $$r\cdot \vec{v}$$ called the scalar product. These operations have to satisfy the following properties.

Associativity: $$(\vec{u}+\vec{v})+\vec{w}=\vec{u}+(\vec{v}+\vec{w})$$

Commutativity: $$\vec{v}+\vec{w}=\vec{w}+\vec{v}$$

Zero element: there is a vector $$\vec{o}$$ such that $$\vec{o}+\vec{v}=\vec{v}$$ and $$\vec{v}+\vec{o}=\vec{v}$$ for every vector $$\vec{v}$$

Negative element: for every $$\vec{v}$$ there is a vector $$-\vec{v}$$ such that $$\vec{v}+(-\vec{v})=\vec{o}$$ and $$(-\vec{v})+\vec{v}=\vec{o}$$

Distributivity (1): $$r\cdot (\vec{v}+\vec{w})=r\cdot \vec{v}+ r\cdot \vec{w}$$

Distributivity (2): $$(r+ s)\cdot \vec{v}=r\cdot \vec{v}+ s\cdot \vec{v}$$

Scalar compatibility: $$(r\cdot s)\cdot \vec{v}=r\cdot (s\cdot \vec{v})$$

Identity: $$1\cdot \vec{v}=\vec{v}$$

A very important example of a vector space is the $$2$$-dimensional plane $$V=\mathbb{R}^2$$ as it was discussed in this chapter.

This page titled 22.2: Operations on 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.