# How to Find the Scalar Product of Two Vectors (2D)

The scalar product or dot product is one of the most important mathematical operations related to vectors, most of all because it shows whether two vectors are perpendicular ($90$° between them) or not. When two vectors are perpendicular, they are said to be orthogonal. The rule looks like this:

Rule

### OrthogonalVectors

 $\stackrel{\to }{a}\cdot \stackrel{\to }{b}=0⇔\stackrel{\to }{a}\perp \stackrel{\to }{b}$

You have an equivalence arrow between the expressions. This means that if one of them is true, the other one is also true.

There are two formulas for finding the dot product (scalar product). One is for when you have the coordinate form of the two vectors, and the other is used when you know the length of the vectors and the angle between them.

Formula

### TheScalarProduct

$\begin{array}{llll}\hfill \stackrel{\to }{u}\cdot \stackrel{\to }{v}& =\phantom{\rule{-0.17em}{0ex}}\left({x}_{1},{y}_{1}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left({x}_{2},{y}_{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}_{1}{x}_{2}+{y}_{1}{y}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \stackrel{\to }{u}\cdot \stackrel{\to }{v}& =\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|\cdot \mathrm{cos}\alpha ,\phantom{\rule{1em}{0ex}}\alpha =\angle \phantom{\rule{-0.17em}{0ex}}\left(\stackrel{\to }{u},\stackrel{\to }{v}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \stackrel{\to }{u}\cdot \stackrel{\to }{v}& =\phantom{\rule{-0.17em}{0ex}}\left({x}_{1},{y}_{1}\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left({x}_{2},{y}_{2}\right)={x}_{1}{x}_{2}+{y}_{1}{y}_{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \stackrel{\to }{u}\cdot \stackrel{\to }{v}& =\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|\cdot \mathrm{cos}\alpha ,\phantom{\rule{2em}{0ex}}\alpha =\angle \phantom{\rule{-0.17em}{0ex}}\left(\stackrel{\to }{u},\stackrel{\to }{v}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 1

Decide whether the vectors $\phantom{\rule{-0.17em}{0ex}}\left(-4,5\right)$ and $\phantom{\rule{-0.17em}{0ex}}\left(2,3\right)$ are orthogonal.

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(-4,5\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(2,3\right)& =-4\cdot 2+5\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-8+15\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =7\ne 0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\phantom{\rule{-0.17em}{0ex}}\left(-4,5\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(2,3\right)=-4\cdot 2+5\cdot 3=-8+15=7\ne 0$

Because the dot product is not equal to zero, you know that the vectors are not orthogonal.

Example 2

Find the scalar product of the vectors $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$, when $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|=3$, $\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|=5$, and the angle between them is $\alpha =\text{}90\text{}\text{°}$.

$\begin{array}{llll}\hfill \stackrel{\to }{u}\cdot \stackrel{\to }{v}& =\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|\cdot \mathrm{cos}\alpha \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot 5\cdot \mathrm{cos}90\text{°}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot 5\cdot 0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Because the scalar product is $0$, you know that $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ are orthogonal.

Example 3

Find $t$ such that $\phantom{\rule{-0.17em}{0ex}}\left(-2,5\right)$ and $\phantom{\rule{-0.17em}{0ex}}\left(2t,9\right)$ are orthogonal.

For two vectors to be orthogonal, the dot product must be equal to $0$. $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(-2,5\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(2t,9\right)& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -4t+45& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill t& =\frac{45}{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

For $t=\frac{45}{4}$, the vectors are orthogonal. In that case, the second vector is

 $\phantom{\rule{-0.17em}{0ex}}\left(2t,9\right)=\phantom{\rule{-0.17em}{0ex}}\left(\frac{45}{2},9\right).$