Math 210 Notes on Cross Products This page requires Mozilla/Firefox/Netscape
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Math 210 Calculus III
The Cross Product: An Alternative Geometric
Approach
[work in progress]
A. Review Dot Product in R^{3} 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 R^{3}. 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 R^{3,} 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, andk×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'.
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.
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: