Math 210 Notes on Cross Products This page requires Mozilla/Firefox/Netscape 7+ or Internet Explorer 6+ MathPlayer.


Math 210 Calculus III
 The Cross Product: An Alternative Geometric Approach
[work in progress]

A. Review Dot Product in R3 and relation to orthogonal vectors.
Recall  the dot product of two vectors in 3 dimensions:
Example: <1,6,2> . <2,1,-3> = 2+6 -6 = 2 = ||<1,6,2>|| ||<2,1,-3|| cos(t) where t is the angle between the two vectors.
Two vectors v and w are orthogonal if and only if v.w = 0.


B. Review of  equations of planes in R3.
Suppose N = <A,B,C> is the normal vector for the plane. So N is orthogonal to every vector in the plane. If (a,b,c) is a point in the plane then
<A,B,C>.<x-a,y-b,z-c>=0 or  A(x-a)+B(y-b)+C(z-c)=0
Example: N = <1,6,2>  and the plane passes through the point (2,1,-3). The equation for the plane is 1 (x-2) + 6 (y-1) + 2 (z+3) = 0.

C. Three problems:
Problem 1. Given two vectors, find a vector that is normal to the plane through the origin that they determine.
Solution example: Suppose v=<2,1,6>, w=<1,4,2>, and the normal vector is N=<x,y,z>.
Then by the orthogonality of  N we have :

2x +1y +6z = 0
1x +4y+2z=0
For an easy solution... suppose z=1. This gives two equations in the variables x and y:
2x +1y = -6
1x +4y = -2
OR
8x +4y = -24
1x +4y = -2
So
7x = -22 and x = -22/7, y = -6 + 44/7 = 2/7.
So one solution is N = <-22/7, 2/7, 1>.  More generally the normal could be any scalar multiple of this vector: t <-22/7, 2/7, 1>. One easier normal vector would be <-22,2,7> 
Problem 2. Given two vectors in R3, find the area of the parallelogram they determine.

Solution: Let t be the angle between the two vectors [which can be determined using the dot product].
The area of the parallelogram is ||v|| h where h is the altitude of the parallelogram using v for the base.
But the h/||w||=sin(t) so the Area is ||v|| ||w|| sin(t)

Problem 3. Given three non-coplanar vectors, v,w, and z  find the volume of the parallel-piped they determine.
Solution: Use two of the vectors v and w to determine the area, A, of the "base" parallelogram.
Find a vector N that is normal to the plane those two vectors determine. Project the third vector z onto that normal vector to determine the altitude of the solid, h, so h/||z||= cos(r) where r is the acute angle between N and z.
Thus  Volume = (A ) (h) = A ||z|| cos(r) = [ ||v|| ||w|| sin(t)] ||z|| cos(r).

Geometric Definition: v×w is the vector that is
(i) normal to the plane (satisfying the right hand rule)
determined by v and w with
(ii) ||vxw|| = the area of the parallelogram determined by v and w.

Furthermore-- the volume of the parallel-piped determined by the three vectors `v,w,` and `z = | ||v×w|| ||z|| cos(r) | = |(v×w) · z|,` where `r` is the measure of the angle between `z`  and `||v × w||`. 

An Algebraic approach to these problems:

Let's make things much simple as a starter.
Consider the standard vectors, i = <1,0,0> and j=<0,1,0>  then...

||i×j|| = 1
and using the right hand rule for i and j we see that the product is in the same direction as the vector k = <0,0,1>.
Thus i×j = k.

BY changing the order of the factors we see (using the right hand rule again) that   j×i = -k. Considering the area feature of the cross product we can see as well that i×i, j×j = 0, and k×k=0.

We can generalize these remarks to any vectors v and w, namely, v×w = -w×v, ("anti-commutativity" or "alternating")  and v×v = 0.

Notice that if the vector v and w are any vectors,  then the parallelogram  determined by the vector av and w will have area a times the area of the parallelogram determined by v and w, so  ||(av)×w||=|a|||v×w||  and so av×bw = ab (v×w) for any real numbers a and b,


By a similar argument using the  the right hand rule we can see that  j×k = i ,  k×j = -i ,  i×k =-j ,  and  k×i = j.

To extend these formulae to all three dimensional vectors we assume that this product distributes over addition:

that is, (v+v')×w = v×w +v'×w  and v×(w+w') = v×w +v×w'.
This gives the formula:
v×w = <a,b,c>×<p,q,r> = (ai+bj+ck)×(pi+qj+rk)
= aqk-arj -bpk +bri +cpj-cqi
=(br-cq)i - (ar- cp)j +(aq-bp)k
= <br-cq, -(ar-cp), aq-bp>.
Using this formula we need to check
(i) This vector is orthogonal to `<a,b,c>` and ` <p,q,r>`.
Check!
`<br-cq, -(ar-cp), aq-bp><a,b,c> = 0` Check:`a(br-cq) -b(ar-cp)+c(aq-bp) = 0` 
<br-cq, -(ar-cp), aq-bp><p,q,r> = 0

(ii) The magnitude of this vector is the area of the parallelogram determined by <a,b,c> and <p,q,r>.
Check! Consider ` ||v×w||^2 = (br-cq)^2 + (ar-cp)^2 + ( aq-bp)^2 = .... = ||v||^2  ||w ||^2 sin(t)`,  where ` t` is the angle between `v` and `w`.
(iii) The vector satisfies the right hand rule on the vectors i, j , and k.  [ and in fact for all vectors!]


Methods for computing the cross product and the volumes of parallel-piped using "determinants".
(Work in progress.)

The determinant  of  two 2 dimensional vectors:
Definition: DET(<a,b>,<c,d>) = ad - bc.
Notice that with this definition
DET(<a,b>,<p,q>) = aq - bp;
DET(<a,c>,<p,r>) = ar - cp; and
DET(<b,c>,<q,r>) = br - cq.
Thus v×w
= <a,b,c>×<p,q,r>
= <br-cq, -(ar-cp), aq-bp>
=<DET(<b,c>,<q,r>),-DET(<a,c>,<p,r>),DET(<a,b>,<p,q>)> 
 =(ai+bj+ck)×(pi+qj+rk)
= aqk-arj -bpk +bri +cpj-cqi
=(br-cq)i - (ar- cp)j +(aq-bp)k
=DET(<b,c>,<q,r>)i - DET(<a,c>,<p,r>)j +DET(<a,b>,<p,q>)k

The two 2 dimensional vectors `<a,b>` and `<b,c>` can be arranged in two rows of an array or matrix

`((a,b),(c,d))`
and then the determinant of this matrix is defined by:
`DET((a,b),(c,d)) =|(a,b),(c,d)| = ad-bc`

The determinant of three 3 dimensional vectors:
 Definition: DET(<a1,b1,c1>,<a2,b2,c2>,<a3,b3,c3>) =
a1 Det(<b2,c2> <b3,c3>) - b1Det(<a2,c2><a3,c3>) + c1Det(<a2,b2>,<a3,b3>).
`DET((a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3))= a_1 DET((b_2,c_2) ,(b_3,c_3)) - b_1DET((a_2,c_2),(a_3,c_3)) + c_1DET((a_2,b_2),(a_3,b_3))`
OR
`|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| = a_1 |(b_2,c_2) ,(b_3,c_3)| - b_1|(a_2,c_2),(a_3,c_3)| + c_1|(a_2,b_2),(a_3,b_3)|`
Examples and shortcuts, Connection to v×w.

Problem 1 Example revisited: Given two vectors, `v=<2,1,6>, w=<1,4,2>`, find a vector that is normal to the plane through the origin determined by `v` and `w.
Since the cross product is normal to the plane, the solution requires only the computation of the cross product
`v times w = <2,1,6> times <1,4,2> = <br-cq, -(ar-cp), aq-bp> `
                                                        `= <1*2-6*4, -(2*2-6*1), 2*4-1*1> = <-22, +2, 7>`.
This confirms the result we obtained previously by solving equations.

Problem 2 revisited: Find the area of the parallelogram determined by two vectors, `v` and `w` we need only  find `| | v times w| |`.
For `v=<2,1,6>, w=<1,4,2>` the area is `||<-22, 2, 7>|| = sqrt{ 484 +4+49} = sqrt{537}.`

Problem 3 revisited:
Using the determinant matrix notation it should be noted that
`|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| = (a_1,b_1,c_1) cdot ( (a_2,b_2,c_2) times (a_3,b_3,c_3)) `
so the absolute value of this determinant is the volume of the parallel-piped determined by the three vectors.
Problem: To find the area of a parallelogram determined by two vector in the plane. 

Solution: Suppose the two vectors are `<a,b>` and `<c,d>`.
Consider the vectors in space `<a,b,0>` and `<c,d,0>`.
Then consider the parallel-piped formed with these two vectors together with the vector `<0,0,1>`,  which will have volume equal in magnitude to the area of the parallelogram.
But this volume is `|<0,0,1> cdot (<a,b,0> times <c,d,0>)|`
`= | |(0,0,1),(a,b,0),(c,d,0)| | = |0 cdot DET((b,0) ,(d,0)) - 0 cdot DET ((a,0),(c,0)) + 1 cdot DET ((a,b),(c,d))|= |ad-bc|. `
Using i, j, and k with the determinant to compute  `<a,b,c>times<p,q,r>`:
Recall that `<a,b,c>times<p,q,r> = DET((b,c),(q,r))i - DET((a,c),(p,r))j +DET((a,b),(p,q))k `
If we treat i, j, and k as numbers - this is
`<a,b,c>times<p,q,r> =DET ((i,j,k),(a,b,c),(p,q,r))=|(i,j,k),(a,b,c),(p,q,r)|`.
Some use this to remember how to compute the cross product by remembering a systematic way to compute this determinant.
Ways to remember how to compute the determinant of a 3 by 3 matrix:
Recall that `DET((a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3))= a_1 DET((b_2,c_2) ,(b_3,c_3)) - b_1DET((a_2,c_2),(a_3,c_3)) + c_1DET((a_2,b_2),(a_3,b_3))`
OR
`|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| = a_1 |(b_2,c_2) ,(b_3,c_3)| - b_1|(a_2,c_2),(a_3,c_3)| + c_1|(a_2,b_2),(a_3,b_3)|`
Another way to organize the computation of this determinant is to recognize that the determinant involves a sum of six terms, each term involves exactly one of the letters a,b, or c and one of the subscripts 1,2, or 3. Half of the coefficients are + and half are -.
One way that people keep this straight is to have a 3 by 5 matrix which repeats the first two columns in the 4th and 5th columns:
`((a_1,b_1,c_1,a_1,b_1),(a_2,b_2,c_2,a_2,b_2),(a_3,b_3,c_3,a_3,b_3))`

Reading down the three diagonals from left to right gives the three positive summands:
`a_1b_2 c_3,  b_1 c_2 a_3,  c_1 a_2 b_3`.

Reading down the three diagonals from right to left gives the three negative summands.

`-c_1b_2 a_3, -a_1 c_2 b_3,  - b_1 a_2 c_3`.