The scalar product of two vectors

You are already familiar with one kind of product that involves vectors. It is the multiplication of a scalar with a vector, and the result is a new (stretched) vector:

Now we introduce a multiplication between two vectors, and the result is a scalar.

Consider two vectors $\vec{a}=\left(\begin{array}{r} a_x\\ a_y\\ a_z \end{array}\right)$ and $\vec{b}=\left(\begin{array}{r} b_x\\ b_y\\ b_z \end{array}\right)$. The scalar product of $\vec{a}$ and $\vec{b}$, written $\vec{a} \bullet \vec{b}$ (or $\vec{a} \cdot \vec{b}$), is defined by multiplying component-wise, and then adding the three resulting products to form a single number:

So we take two vectors (six numbers), and by forming the scalar product we obtain one number!

Note that in the second example we have $\vec{a}\neq 0$ and $\vec{b}\neq 0$ but $\vec{a}\bullet \vec{b} =0$. This is a key difference to the normal product between two numbers. If we have two numbers $a\neq 0$ and $b\neq 0$ it always follows $a\cdot b\neq 0$. However, many other properties are similar, e.g.

1. $\vec{a}\bullet\vec{a} = \vert\vec{a}\vert^2 \geq 0$ (squares are not negative)
2. $\vec{a}\bullet\vec{b} = \vec{b}\bullet\vec{a}$ (product is commutative)
3. $(c \cdot \vec{a})\bullet \vec{b} =c\cdot (\vec{a}\bullet \vec{b})$ where $c$ is s scalar (product is associative, that is, you can group)
4. $\vec{a}\bullet (\vec{b}+\vec{c}) = (\vec{a}\bullet\vec{b}) + (\vec{a}\bullet\vec{c})$ (product is distributive, that is, the product distributes over the sum)