What is the dot product and why do we need it? I understand how to calculate the dot product of the vectors. But I don't actually understand what a dot product is, and why it's needed.
Could you answer these questions?
 A: Dot products are very geometric objects. They actually encode relative information about vectors, specifically they tell us "how much" one vector is in the direction of another. Particularly, the dot product can tell us if two vectors are (anti)parallel or if they are perpendicular.
We have the formula $\vec{a}\cdot\vec{b} = \lVert \vec{a}\rVert\lVert \vec{b}\rVert\cos(\theta)$, where $\theta$ is the angle between the two vectors in the plane that they make. If they are perpendicular, $\theta = 90^{\circ}, 270^{\circ}$ so that $\cos(\theta) = 0$. This tells us that the dot product is zero. This reasoning works in the opposite direction: if the dot product is zero, the vectors are perpendicular.
This gives us a quick way to tell if two vectors are perpendicular. It also gives easy ways to do projections and the like.
A: That's a huge question. It's what's called an inner product. A good short answer is that it gives you a way to make sense of what an angle between two vectors is. $$\theta = \cos^{-1}\left(  \frac{a\cdot b }{|a||b|} \right)$$
A: In a general vector spaces you can define the length of a vector by the induced norm via 
$$\|x\| = \sqrt{x\cdot x}$$
this is possible because the dot product is positive definite and thus $x\cdot x$ is not-negative. 
It is even possible to define an angle between to vectors this way by
$$\phi = arccos \frac{x\cdot y}{\sqrt{x\cdot x}\cdot\sqrt{y \cdot y}} $$
also two vector are orthogonal iff their inner product is zero, i.e.
$$ x \perp y \Longleftrightarrow x\cdot y = 0$$
Note that this is possbile for every vector space that has an inner product (dot product) 

A more special example could be: Take the vector space of the continous functions on the intervall $\left[-1,1\right]$ with the inner product defined by $\int_{-1}^1 f(x)g(x) dx$, then the functions $f(x)=x$ and $g(x)=x^2$ are orthogonal, because
$$ \int_{-1}^{1}x\cdot x^2 dx = \int_{-1}^{1} x^3 dx = 0$$
And the length of $f$ would be
$$\left\|f\right\| = \sqrt{\int_{-1}^{1}x\cdot x\, dx } = \sqrt{\frac{2}{3}}$$ 
A: Consider two points in $\mathbb R^n$, $P=(x_1, x_2, \dots, x_m)$ and $Q=(y_1, y_2, \dots, y_n)$. Let $O=(0,0,\dots, 0)$ be the origin. How do you find $\theta = m\angle POQ$? According to the law of cosines, 
$$\cos \theta = \dfrac{|P|^2 +|Q|^2-|Q-P|^2}{2|P|\cdot|Q|}$$.
$$|P|^2=\sum_{i=1}^n x_i^2$$ 
$$|Q|^2=\sum_{i=1}^n y_i^2$$
$$|Q-P|^2=\sum_{i=1}^n (y_i-x_i)^2
        =\sum_{i=1}^n (x_i^2+2x_iy_i+y_i^2)$$
It follows that $\displaystyle |P|^2 +|Q|^2-|Q-P|^2 = 2\sum_{i=1}^n x_iy_i$
So
$$\cos \theta = \frac{\sum_{i=1}^n x_iy_i}{|P|\cdot|Q|}$$.
It is convenient to define the dot product
$P \circ Q = \sum_{i=1}^n x_iy_i$.

So now consider an n-dimensional vector space $\mathbf V$ with a basis 
$\{\mathbf e_i\}_{i=1}^n$. Given $P=\sum_{i-1}^n x_i \mathbf e_i$ and $Q=\sum_{i-1}^n y_i \mathbf e_i$ we define $P \circ Q = \sum_{i=1}^n x_iy_i$ and we define $P$ is orthogonal to $Q$ when $P \circ Q = 0$.
We can now define the angle, $\theta$ between the vectors $P$ and $Q$ by 
$\cos \theta = \dfrac{P \circ Q}{|P| \cdot |Q|}$.
Thus the dot product is just an abstraction of a property of vectors in the Euclidean plane. From here we can now define what an orthogonal and an orthonormal basis are and find some very interesting consequences. Yes, there are a lot of technical problems that need to be solved, but they can be solved.
A: If the length of $B$ is $1$ then $\langle A,B\rangle$ is the coordinate of $A$ in direction $B$.
There is a nice interpretation of the scalar product where $B$ has arbitrary length.  Let $B=(b_1,b_2)$, then define $J(B):=(-b_2,b_1)$; you'll get $J(B)$ by rotating $B$ counterclockwise by $\pi/2$.  Observe that
$$\langle A, B\rangle=\det\bigl(A,J(B)\bigr),$$
that is: the dot product is the (orientated) area of the parallelogram spanned by $A$ and $J(B)$.
