Scalar or dot product

There are two ways to multiply two vectors. Here, we will learn about the scalar product. It has many applications in STEM. For example, scalar products are used to calculate the work done by a system, in computer graphics to calculate the amount of light hitting surfaces, and in engineering to determine the stress applied to materials.

The scalar product is the result of multiplying the magnitudes of the components of two or more vectors. This does not result in a vector, but a scalar; a scalar or dot product does not have a direction.

For two vectors $\overrightarrow{a}(a_1,a_2,a_3)$ and $\overrightarrow{b}(b_1,b_2,b_3)$. The scalar product is defined by:

$$\overrightarrow{a}\cdot \overrightarrow{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$$

or

$$\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$$ where $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.

It is also commonly called the dot product as a dot is used between the two vectors being multiplied instead of a multiplication symbol.

Properties of the scalar or dot product

There are some key characteristics of scalar products that are handy to remember.

1. If $\overrightarrow{a}$ and $\overrightarrow{b}$ are non-zero vectors and $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$ (i.e. there is ${90}^{\circ }$ or ${\displaystyle \frac{\pi }{2}}$ radians between them), then $\overrightarrow{a}\cdot \overrightarrow{b}=0$, since $\cos \left({90}^{\circ }\right)=\cos \left({\displaystyle \frac{\pi }{2}}\right)=0$.
2. If $\overrightarrow{a}$ is parallel to $\overrightarrow{b}$, then the angle between the vectors is $0$ and $\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|$ as $\cos \left(0\right)=1$.
3. The dot product does not depend on the order of multiplication, so $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}$.
4. In three dimensions, with $\hat{i}$, $\hat{j}$ and $\hat{k}$ unit vectors along the $x$-, $y$- and $z$-axes, respectively: $$\begin{array}{rl}\overrightarrow{i}\cdot \overrightarrow{j}& =\overrightarrow{j}\cdot \overrightarrow{k}=\overrightarrow{k}\cdot \overrightarrow{i}=0\\ \overrightarrow{i}\cdot \overrightarrow{i}& =\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1\end{array}$$

Example 1 – calculating the scalar product

$$\begin{array}{rl}(2\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k})\cdot (-\overrightarrow{i}-2\overrightarrow{j}+\overrightarrow{k})& =(2\times (-1))+(3\times (-2))+(4\times 1)\\ & =-4\end{array}$$

Exercise – calculating the scalar product

1. Calculate the dot products of the following pairs of vectors.
1. $(2,5,-1)$ and $(4,1,1)$
2. $3\overrightarrow{i}$ and $5\overrightarrow{j}$
3. $5\overrightarrow{k}$ and $(\overrightarrow{j}+2\overrightarrow{k})$
2. Find:
1. $(2,0,4)\cdot (-3,1,3)$
2. $(0,5,1)\cdot (4,0,0)$
3. $(2\overrightarrow{i}+3\overrightarrow{k})\cdot (7\overrightarrow{i}+2\overrightarrow{j}+4\overrightarrow{k})$
3. Determine which of the following vectors are perpendicular.
1. $(5,2,3)$
2. $(0,1,-1)$
3. $(-2,2,2)$

Angle between two vectors

The angle $\theta$ where $(0\le \theta \le \pi )$, between two vectors can be found using same equations used to calculate the scalar product:
$$\overrightarrow{a}\cdot \overrightarrow{b}=\mid \overrightarrow{a}\mid \mid \overrightarrow{b}\mid \cos \theta$$
We can rearrange this to make $\theta$ the subject:

$$\begin{array}{rl}\cos \theta & =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\\ \theta & ={\cos }^{-1}\left(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}\right)\end{array}$$

Example 1 – finding the angle between two vectors

If $\overrightarrow{a}=(2,3,1)$ and $\overrightarrow{b}=(5,-2,2)$, find the angle $\theta$ between $\overrightarrow{a}$ and $\overrightarrow{b}$.
We can start by finding $\overrightarrow{a}\cdot \overrightarrow{b}$.
$$\begin{array}{rl}\overrightarrow{a}\cdot \overrightarrow{b}& =(2,3,1)\cdot (5,-2,2)\\ & =6\end{array}$$
Now, let's find $\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$.
$$\begin{array}{rl}\left|\overrightarrow{a}\right|& =\sqrt{2^2+3^2+1^2}\\ & =\sqrt{14}\end{array}$$
$$\begin{array}{rl}\left|\overrightarrow{b}\right|& =\sqrt{25+4+4}\\ & =\sqrt{33}\end{array}$$
Let's substitute these values into the rearranged scalar product equation.
$$\begin{array}{rl}\theta & ={\cos }^{-1}\left(\frac{6}{\sqrt{33}\times \sqrt{14}}\right)\\ & ={\cos }^{-1}\left(0.2791\right)\\ & ={73.8}^{\circ }\end{array}$$
Therefore, the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is ${73.8}^{\circ }$.

Exercise – finding the angle between two vectors

1. Find the angle between the following pairs of vectors.
1. $(1,2,3)$ and $(4,-1,0)$
2. $(2,1,-2)$ and $(1,5,-1)$
3. $(0,5,1)$ and $(2,0,0)$
4. $(1,-2,3)$ and $(-4,1,-3)$
5. $(2,1,-2)$ and $(0,4,0)$
6. $(0,3,0)$ and $(0,1,0)$
2. Consider the vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ where $\overrightarrow{a}=(2,2,2)$, $\overrightarrow{b}=(3,2,-1)$ and $\overrightarrow{c}=(-1,4,1)$.
1. Show that $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{a}\cdot \overrightarrow{c}$.
2. Rearranging $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{a}\cdot \overrightarrow{c}$ gives $\overrightarrow{a}\cdot (\overrightarrow{b}-\overrightarrow{c})=0$. Assuming that $\overrightarrow{b}\ne \overrightarrow{c}$, determine the relationship between $\overrightarrow{a}$ and $(\overrightarrow{b}-\overrightarrow{c})$.

Answers

1. 1. ${82.6}^{\circ }$
   2. ${54.7}^{\circ }$
   3. ${90}^{\circ }$
   4. ${141.8}^{\circ }$
   5. ${70.5}^{\circ }$
   6. $0^{\circ }$
2. 1. $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{a}\cdot \overrightarrow{c}=8$
      $\begin{array}{rl}\overrightarrow{a}\cdot \overrightarrow{b}& =(2,2,2)\cdot (3,2,-1)\\ & =(2\times 3)+(2\times 2)+(2\times (-1))\\ & =8\end{array}$
      $\begin{array}{rl}\overrightarrow{a}\cdot \overrightarrow{c}& =(2,2,2)\cdot (-1,4,1)\\ & =(2\times (-1))+(2\times 4)+(2\times 1)\\ & =8\end{array}$
   2. $\overrightarrow{a}$ is perpendicular to $(\overrightarrow{b}-\overrightarrow{c})$.

Images on this page by RMIT, licensed under CC BY-NC 4.0

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Keywords

• Dynamics
• Maths
• Physics
• Vectors

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