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(Sorry for my very basic knowledge, I don't know much about math :) ).

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? Thanks

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One use is to quantify how 'aligned' two vectors are. In particular, if the dot product of two vectors is zero, they are deemed orthogonal. –  copper.hat Jan 6 at 18:46

4 Answers 4

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Dot products are very geometrical 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. Hopefully this clears things up a bit.

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Good explanation... and it's interesting to note that this interpretation of projection goes far beyond the scope of geometry. You can use (in fact, you do use) this notion on much more abstract vector spaces, like the Hilbert Spaces. The projection here is essential for applications in the statistical interpretation of Quantum Mechanics. Anyway, the important is that dot product measures how much one (abstract) vector is in the direction of the other. –  GabrielF Jan 6 at 23:52

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. $$\theta = \cos^{-1}\left( \frac{a\cdot b }{|a||b|} \right)$$

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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

$$\left\|f\right\| = \sqrt{\int_{-1}^{1}x\cdot x\, dx } = \sqrt{\frac{2}{3}}$$

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If-and-only-if is spelled "iff", not "iif". –  AJMansfield Jan 7 at 2:40
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I'd be surprised if the OP gets any benefit from examples involving vector spaces of functions. –  bubba Jan 7 at 13:09
    
@bubba It's just an additional example to illustrate that this works for every vector space with an inner product. The real answer is clearly seperated above the "-----" –  user127.0.0.1 Jan 7 at 13:15

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 $J(B)$.

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