Mini-Post: Dot Products and Cross Products, and their Equivalent Trig Things.

August 17, 2010

Okay, so, we know (or you should know!) what cross and dot products are for Euclidean spaces.  We use them all the time!  On the daily!  So, I’m sure you’ve seen the equalities

a\cdot b = \|a\|\|b\|\cos(\theta)

\|a\times b\| = \|a\|\|b\|\sin(\theta)

where \theta is the angle between the vectors a and b.  Where do these equalities come from?  Let’s find out.

The Dot Product.


Note that, in the picture above, we have constructed a triangle.  Neat.  Now, if you’re anybody who’s anybody, you are well aware of the Law of Cosines.  Even though I know you can recite it from memory, let’s write it out explicitly below:

x^2 = y^2 + z^2 - 2xy\cos(\theta_{z})

where \theta_{z} is the angle opposite the side z.

Well, we have a triangle here, so let’s apply this.  We get:

(\|a-b\|)^2 = \|a\|^2 + \|b\|^2 - 2\|a\|\|b\|\cos(\theta)

But note that \|x\|^2 = x\cdot x for any x.  That’s convenient.  We can rewrite this equality as

(a-b)\cdot (a-b) = \|a\|^2 + \|b\|^2 - 2\|a\|\|b\|\cos(\theta)

but we can reduce the left side

(a-b)\cdot (a-b) = a\cdot a - a\cdot b - b\cdot a + b\cdot b

= a\cdot a - 2(a\cdot b) + b\cdot b

= \|a\|^2 - 2(a\cdot b) + \|b\|^{2}

and so, we can rewrite this by replacing the left side of the original equation with this one.  We get:

\|a\|^2 - 2(a\cdot b) + \|b\|^{2} = \|a\|^2 + \|b\|^2 - 2\|a\|\|b\|\cos(\theta)

Reducing this, we get

-2(a\cdot b) = -2\|a\|\|b\|\cos(\theta)


a\cdot b = \|a\|\|b\|\cos(\theta)

And this is the equation that we’re used to havin’!  Good.

Cross Product.

For the cross product, we require three dimensions.  So, every vector will be represented by: a = a_{1}i + a_{2}j + a_{3}k, where i,j,k represent the unit vector in the x,y,z direction, respectively.  Note that this is exactly the same as saying a = (a_{1}, a_{2}, a_{3}), but this first way is the usual “physic” notation, and it is slightly easier to work with.  We have the same picture as above in the dot product, and we want to show that

\|a \times b\| = \|a\|\|b\|\sin(\theta)

so, where’s a good place to start?  Well, let’s write everything out.

a\times b = (a_{1}i + a_{2}j + a_{3}k) \times (b_{1}i + b_{2}j + b_{3}k)

Now, let’s distribute.

= a_{1}b_{1}(i\times i) + a_{1}b_{2}(i\times j) + a_{1}b_{3}(i\times k)

+ a_{2}b_{1}(j\times i) + a_{2}b_{2}(j\times j) + a_{2}b_{3}(j\times k)

+ a_{3}b_{1}(k\times i) + a_{3}b_{2}(k\times j) + a_{3}b_{3}(k\times k)

Okay, but what’s i\times i?  If you want to figure it out matrix-wise, you can, but it is just going to turn out to be zero.  We get a lot of things crossing out because of this.  We wind up having the significantly messier expression

= a_{1}b_{2}k - a_{1}b_{3}j - a_{2}b_{1}k + a_{2}b_{3}i + a_{3}b_{1}j - a_{3}b_{2}i

= (a_{2}b_{3} - a_{3}b_{2})i + (a_{3}b_{1} - a_{1}b_{3})j + (a_{1}b_{2} - a_{2}b_{1})k

which is a pretty nice deal, all things considered.

Okay, now, here’s the crazy part.  I’m not going to do everything completely out, but you can write it out on paper if you’d like.  From above, we have

\|a\times b\|^{2} = (a_{2}b_{3} - a_{3}b_{2})^{2} + (a_{3}b_{1} - a_{1}b_{3})^{2} + (a_{1}b_{2} - a_{2}b_{1})^{2}

and this, ugh, is equal to

= a_{2}^{2}b_{3}^{2} - 2a_{2}a_{3}b_{2}b_{3} + a_{3}^{2}b_{2}^{2}

+ a_{3}^{2}b_{1}^{2} - 2a_{1}a_{3}b_{1}b_{3} + a_{1}^{2}b_{3}^{2}

+ a_{1}^{2}b_{2}^{2} - 2a_{1}a_{2}b_{1}b_{2} + a_{1}^{2}b_{2}^{2}

BUT, we can make this much nicer.  This is the part you might wanna scratch out on paper to see that it works:

= (a_{1}^{2} + a_{2}^{2} + a_{3}^{2})(b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) - (a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3})^{2}

= \|a\|^{2}\|b\|^{2} - (a\cdot b)^{2}

= \|a\|^{2}\|b\|^{2} - \|a\|^{2}\|b\|^{2}\cos^{2}(\theta)

from the dot product identity above, which gives us, finally:

= \|a\|^{2}\|b\|^{2}(1 - \cos^{2}(\theta))

= \|a\|^{2}\|b\|^{2}(\sin^{2}(\theta))

and putting this all together, we get

\|a\times b\|^{2} = \|a\|^{2}\|b\|^{2}(\sin^{2}(\theta))

which, after taking the square root of both sides (since the norm is always positive, we can do such a thing…), reduces to the significantly nicer:

\|a\times b\| = \|a\|\|b\|\sin(\theta)

which is what we’re used to.

Last Note.

I learned a kind of cute way to remember which product is cosine and which is sine; unfortunately, I don’t remember who taught it to me, so they’re gonna get absolutely no credit for it!  Sad.  Either way, I always say to myself:

{\bf \mbox{Cross ain't Cos}}

which means that “cross product isn’t the cosine one…that’s the dot product.”  If you have some other jingle or way to remember it, comment it!

4 Responses to “Mini-Post: Dot Products and Cross Products, and their Equivalent Trig Things.”

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