Dot product of vectors

[jpanknin]

The following shows two examples of vector multiplication to get the dot product. It's easy enough to understand the math, but what use/value is there for the single number that results? What does the answer (2 in the first example and the 4 in the second example) represent? I know it's used later to find the angle between the two vectors, but is there any application of the dot product itself? Can we geometrically or algebraically use the '2' and the '4' dot products to tell us anything else about the vectors? When you add, subtract, or multiply a vector by a scalar, the vector is transformed somehow. But I'm having trouble understanding the meaning of the dot product. Any help would be appreciated.

 

[ISTER_REG]

So what is the dot product? Here one multiplies component-wise the vectors and receives at the end a scalar (simplified said a number). E.g.:
[MATH]\vec{a} \circ \vec{b} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \circ \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = a_1b_1 + a_2b_2 + a_3b_3[/MATH]This number can indicate e.g. whether two vectors are orthogonal (90 degree angle), then this is 0. Besides the "normal" dot product, there exists also a standard dot product for complex numbers, important is here e.g. the conjugate complex!

 

[Dr.Peterson]

The various uses, generally connected to the angle as you mention, are enough; we could say that the dot product is valuable the way certain chemicals are valuable, as an intermediate material from which other things can be made, while being very easy to make itself.

The simplest description I know of for the dot product itself is that it is the product of the projection of either vector on the other, times the magnitude of the other. So it is just a product of the two magnitudes, taking into account the angle between them.

 

[Zermelo]

For this exact reason, in beginner's Linear Algebra class, we defined the dot product a little differently (in R^2), the way Dr Peterson described. If a and b are vectors, [MATH]a \cdot b = \lvert a \rvert \lvert b \rvert \cos{(\angle (a,b)) }[/MATH]. Aka, the dot product is the product of their intensity and the cosine of the angle between them. Now you can use the dot product to figure out the projection of one vector to another, to find the angle between them etc etc. What makes this great is that you can prove that these definitions are equivalent, and that the dot product is really easy to compute! So if you know what the vectors are, it would really be a pain to compute the angle between them "by hand". But using the dot product, we can figure out [MATH]\lvert a \rvert = \sqrt{a \cdot a}[/MATH], and when we find the vector's intensities, we can easili solve for the angle between them. Also, watch this video (I recommend the whole series actually) for a visual intuition:

 

[lex]

 

[HallsofIvy]

Again, \(\displaystyle u\cdot v= |u||v| cos(\theta)\) where \(\displaystyle |u|\) and \(\displaystyle |v|\) are the lengths of the vectors u and v (Zermelo use the word "intensity" which would certainly apply if these were force vectors or light vectors.) and \(\displaystyle \theta\) is the angle between two angles. One important application of the dot product concerns the fact that if \(\displaystyle \theta\) is 90 degrees or \(\displaystyle \pi/2\) radians (i.e. if u and v are perpendicular) then \(\displaystyle cos(\theta)= 0\). Two vectors, u and v, are perpendicular if and only if \(\displaystyle u\cdot v= 0\).

 

[pka]

Suppose that \(\vec{a}~\&~\vec{b}\) are two vectors which are neither perpendicular nor parallel (i.e.\(\;\;\vec{a}\cdot\vec{b}\ne0~\&~\vec{a}\ne\alpha\vec{b}\))
In that case, the vector \(\vec{a}\) is equal to the sum of a vector parallel to \(\vec{b},\;\vec{b}_{\|}\) and a vector perpendicular to \(\vec{b},\;\;\vec{b}_{\bot }\).
They are \(\vec{b}_{\|}=\frac{\vec{a}\cdot\vec{b}}{\vec{b}\cdot\vec{b}}\vec{b}~\&~\vec{b}_{\bot }=\vec{b}-\vec{b}_{\| } \)