# Scalar Product

While adding vectors to each other and scaling a vector by multiplication by a real number have natural interpretations, we shall now introduce a (first) product of two vectors that is less motivated at first sight,  but in fact is very fundamental and useful.

Given two vectors $a=(a_1,a_2)\in R^2$ and $b=(b_1,b_2)\in R^2$, we define their scalar product $a\cdot b$ by

• $a\cdot b=a_1b_1+a_2b_2$.

We note, as the terminology suggests, that the scalar product $a\cdot b$ of two vectors$a$ and $b$ is a scalar, that is a number in $R$, while the factors $a$ and $b$ are vectors in $R^2$. Note also that forming the scalar product of two vectors involves not only multiplication, but also a summation!

We note the following connection between the scalar product $a\cdot a$ and the norm $\vert a\vert$:

• $a\cdot a=\vert a\vert^2=a_1^2+a_2^2$,
• $\vert a\vert =\sqrt{a\cdot a}=(a\cdot a)^{\frac{1}{2}}$.

Below we shall define another type of product of vectors where also the product is a vector.

We shall thus consider two different types of products of two vectors, which we will refer to as the scalar product and the vector product, respectively.

At first when limiting our study to vectors in $R^2$, we may also view the vector product to be a single real number. However, the vector product in $R^3$ is indeed a vector in $R^3$.

# Scalar Product as a Function

We may view the scalar product as a function $f:R^2\times R^2\rightarrow R$ where $f(a,b)=a\cdot b$.

To each pair of vectors $a\in R^2$ and $b\in R^2$, we associate the number $f(a,b)=a\cdot b\in R$.

Similarly we may view summation of two vectors as a function $f:R^2\times R^2\rightarrow R^2$.

Here, $R^2\times R^2$ denotes the set of all ordered pairs $(a,b)$ of vectors $a=(a_1,a_2)$ and $b=(b_1,b_2)$ in $R^2$.

Example: We have $(3,7)\cdot (5,2)=15+14=29$, and $\latex (3,7)\cdot (3,7)=9+49=58$ so that $\vert (3,7)\vert =\sqrt{58}$.

# Properties of the Scalar Product

The scalar product $a\cdot b$  is linear in each of the arguments $a$ and $b$, that is

• $a\cdot (b+c)=a\cdot b+a\cdot c$,
• $(a+b)\cdot c=a\cdot c+b\cdot c$,
• $(\lambda a)\cdot b=\lambda a\cdot b, \quad a\cdot (\lambda b)=\lambda a\cdot b$ ,

for all $a,b\in R^2$ and $\lambda\in R$. This follows directly from the definition. For example, we have

• $a\cdot (b+c)=a_1(b_1+c_1)+a_2(b_2+c_2)$
• $=a_1b_1 + a_2b_2 +a_1c_1 + a_2c_2= a\cdot b+a\cdot c$.

Using the notation $f(a,b)=a\cdot b$, the linearity properties may be expressed as

• $f(a,b+c)=f(a,b)+f(a,c),quad\quad f(a+b,c)=f(a,c)+f(b,c)$,
• $f(\lambda a,b)=\lambda f(a,b)\quad\quad f(a,\lambda b)=\lambda f(a,b)$.

# Scalar Product as Bilinear Form

We also say that the scalar product $a\cdot b=f(a,b)$ is a bilinear form on $R^2\times R^2$, that is a function from $R^2\times R^2$ to $R$, since $a\cdot b=f(a,b)$ is a real number for each pair of vectors $a$ and $b$ in $R^2$ and $a\cdot b=f(a,b)$ is linear both in the variable (or argument) $\latex a$ and the variable $\latex b$.

Furthermore, the scalar product $a\cdot b=f(a,b)$ is symmetric in the sense that

• $a\cdot b=bcdot a\quad\mbox{or}\quad f(a,b)=f(b,a)$,

and positive definite in the sense that

• $a\cdot a=|a|^2>0\quad \mbox{for }a\neq 0=(0,0)$.

We may summarize by saying:

• The scalar product $a\cdot b$ is a bilinear symmetric positive definite form on $R^2\ times R^2$ (with values in $R$.)

We notice that for the basis vectors $e_1=(1,0)$ and $e_2=(0,1)$, we have

• $e_1\cdot e_2=0, \quad e_1\cdot e_1=1,\quad e_2\cdot e_2=1$.

Using these relations, we can compute the scalar product of two arbitrary vectors $a=(a_1, a_2)$ and $b=(b_1,b_2)$ in $R^2$ using the linearity as follows:

• $a\cdot b=(a_1e_1+a_2e_2)\cdot (b_1e_1+b_2e_2)$
• $= a_1b_1e_1\cdot e_1+a_1b_2e_1\cdot e_2$ $+\, a_2b_1e_2\cdot e_1+a_2b_2e_2\cdot e_2$
• $= a_1b_1+a_2b_2$.

We may thus define the scalar product by its action on the basis vectors and then extend it to arbitrary vectors using the linearity in each variable.

# Scalar Product from Projection

We shall now prove that the scalar product $a\cdot b$ of two vectors $a$ and $b$ in $R^2$ can be expressed as

• $a\cdot b=\vert a\vert \vert b\vert\cos(\theta )$,     (1)

where $\theta$ is the angle between the vectors $a$ and $b$.

This formula has a geometric interpretation: Assuming that $\vert \theta\vert le 90$ degrees so that $\cos(\theta )$ is positive, consider the right-angled triangle OAC shown in the figure below.

The length of the side OC is $\vert a\vert \cos(\theta )$ and thus $a\cdot b$ is equal to the product of the lengths of sides OC and OB, if (1) is correct which remains to be shown.

We will refer to OC as the projection of OA onto OB, considered as vectors, and thuswe may say that $a\cdot b$ is equal to the product of the length of the projection of OA onto OB and the length of OB. Because of the symmetry, we may also relate $a\cdot b$ to the projection of OB onto OA, and conclude that $a\cdot b$ is also equal to the product of the length of the projection of OB onto OA and the length of OA:

$a\cdot b=\vert a\vert,\vert b\vert\cos(\theta)$.

To prove (1), we write using the polar representation

• $a=(a_1,a_2)=\vert a\vert (\cos(\alpha ),\sin(\alpha ))$,
• $b=(b_1,b_2)=\vert b\vert (\cos(\beta ),\sin(\beta ))$,

where $\alpha$ is the angle of the direction of $a$ and $\beta$ is the angle of direction of $b$. Using a basic trigonometric formula (to be proved in Trigonometric Functions), we see that

• $a\cdot b=a_1b_1+a_2b_2$
• $=\vert a\vert \vert b\vert (\cos(\alpha )\cos(\beta )$ $+\, sin(\alpha )sin(\beta ))$,
• $=\vert a\vert \vert b\vert\cos(\alpha -\beta )$ $=\vert a\vert\vert b\vert\cos(\theta )$,

where $\theta =\alpha -\beta$ is the angle between $a$ and $b$.

Note that since $\cos(\theta )=cos(-\theta )$, we may compute the angle between $a$ and $b$ as $\alpha -\beta$ or $\beta -\alpha$.

We may thus view $\vert b\vert\vert \cos(\theta )\vert$ as the length of the projection of the vector $b$ in the direction of $a$, as shown in this figure:

Projection of $b$ in the direction of $a$ denoted $P_ab$ of length $\vert b\vert\cos(\theta )$

We will come back to the important concept of projection in more detail below.

Map projection principle.