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Most proofs of the magnitude of the cross product are algebraic in nature, I find I learn best visually / geometrically.

Is there a breakdown of the proof of the magnitude of the cross product using only geometry?

I don't have a (firm) background in linear algebra, so talk of the determinant or matrices isn't going to be terribly helpful for me. =/

I can establish some relationships with the triangles formed by the vectors and their cross product, but I don't want to butt my head against this problem, when I have so many chapters left to complete.

I'm working my way through Calculus 3 this summer to get ready for class in the fall.

Thanks.

Edit: (My longer response to Brian's solution below)

Just to make sure I'm on the same page here:

  • The notation for $a_i$ is the same as the notation my book uses for $a_1$, the portion of the $\vec{a}$ vector in the x direction.
  • The "$\hat{i}$" is the unit vector for the x direction.
  • Your "2D cross product" is the determinant of a 2x2 matrix (from the look of it); I was able to make that leap thanks to your definition in the comments.
  • The "difference of two rectangles" would be the difference of the area of the rectangle formed by $a_i$*$b_j$ (the area of the first) and $a_j$*$b_i$ (second); The "$(a_i b_j - a_j b_i)$" portion.

Given all of that, your solution makes perfect sense. The leap from the book's definition and seeing the $a_i$*$b_j$ as the area of a rectangle is what really made the difference.

The only difference between this solution and the normal cross product, would be adding an extra dimension and the difference of two more rectangles.

Edit2:

And we can test for perpendicular vectors, because - when we construct our difference of rectangles - if the rectangles are equal (e.g. the vectors are perpendicular) then when we subtract the areas we get a sum of zero in each direction.

mathFromtheGroundUp
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  • How do you define the cross product? –  May 29 '15 at 20:56
  • a = a1i + a2j + a3z. b = b1i + b2j + b3z. a x b = (a2b3 - a3b2)i + (a3b1 - a1b3)j + (a1b2 - a2b1)k. Where 'a' and 'b' are vectors, with the components as shown. – mathFromtheGroundUp May 29 '15 at 20:59
  • The magnitude of the cross product of two vectors $\vec{a}$ and $\vec{b}$ is the area of the parallellogram spanned by those two vectors. – Steven Van Geluwe May 29 '15 at 21:01
  • And you want to prove that the norm of this is $|a||b|\sin(\theta)$? That seems entirely algebraic to me. Why don't you consider the geometric definition of the cross product -- that is define the cross product to be the area of the parallelogram determined by $a$ and $b$ with the direction given by the right hand rule? –  May 29 '15 at 21:02
  • Right, that's the basic definition. But not why those two are equal or how one would show it with geometry. @StevenVanGeluwe – mathFromtheGroundUp May 29 '15 at 21:02
  • @Bye_World I was just taking the definition from the book. I'd prefer the geometric definition, when discussing it. It wasn't clear to me which you wanted. =/ And I would like to prove the magnitude of the cross product is the ∥a∥∥b∥sin(θ) as you mentioned, yes. – mathFromtheGroundUp May 29 '15 at 21:05
  • So did I answer your answer below or would you also like a proof that $|a||b|\sin(\theta) = (a_2b_3-a_3b_2)i + \cdots$? –  May 29 '15 at 21:07
  • how about starting with two vectors$u$ and $v$, at right angles to each other, and place the vector $n$ representing the cross product perpendicular to both. Then, using your geometric definition of cross product, this vector will have length equal to the area of the rectangle. Now, see what happens to the area of the base as you move the two original vectors together by decreasing the angle between them. The cross product vector will shrink because the area of the base decreases. Now, how is this decrease related to the angle between $u$ and $v$? – Matematleta May 29 '15 at 21:10
  • There are three properties in the discussion: (a) The cross-product is defined as $\det [\mathbb{1}, a, b]$, where each of those is a three-element column vector. (b) The cross-product has magnitude $|a| |b| \sin \theta$. (c) The cross-product has magnitude equal to the area of a parallelogram "subtended" by $a$ and $b$. Now, which of those three did you want to show equal to each other? – Brian Tung May 29 '15 at 21:13
  • @Chilango I don't get why the area of the rectangle would by the magnitude of n, that's really my basic question. I'm looking for a proof of how they would relate. – mathFromtheGroundUp May 29 '15 at 21:14
  • So you want a proof that the cross product defined by $(a_2b_3-a_3b_2)i + \cdots$ has a magnitude given by the area of a parallelogram? –  May 29 '15 at 21:15
  • @BrianTung, as above, part A of your discussion isn't helpful to me. I don't have background in linear algebra (det, matrices, etc). Part B and C seem to be related, but I lean toward wanting to understand part C (I believe). – mathFromtheGroundUp May 29 '15 at 21:16
  • @Bye_World, geometrically if possible, yes. I believe so. Without your background, I think it's hard to ask the right question. XD – mathFromtheGroundUp May 29 '15 at 21:16
  • Part (a) was my shorthand for your definition of the cross product way up in the second comment. – Brian Tung May 29 '15 at 21:17
  • @BrianTung, I'll have to take your word for it. =/ Thanks for summing that up, I think that should be helpful if I need to discuss the problem in different terms than I'm comfortable with. – mathFromtheGroundUp May 29 '15 at 21:19
  • So you want to go from your algebraic definition of the cross-product, to the magnitude of the cross-product representing the area of the parallelogram? – Brian Tung May 29 '15 at 21:23
  • @BrianTung, I believe so? I think I'll take a page from Bye_World's book and say: Define a cross product to be a perpendicular vector (c) to two other vectors a and b. Given just that information, can we arrive at the magnitude of c = ||a|| ||b|| sin(θ) using just geometrical figures (triangles, parallelograms, squares, etc) to show a proof. – mathFromtheGroundUp May 29 '15 at 21:26
  • @ mathFromtheGroundUp: you just DEFINE the vector perpendicular to the other two to have magnitude equal to the area. Once you do that, then themagnitude varies with $\theta $ as in the formula. Or maybe I do not understand what you mean by a purely geometric definition? – Matematleta May 29 '15 at 21:26
  • @BrianTung, hmm both you and Bye_World have said that "just define the magnitude to be equal to the area of a parallelogram". There's obviously some deficiency in my math background, forgive me. How are you able to make an assertion about the magnitude (a length) being an area? I'll try to work up some visual aides to help explain what I have in mind. – mathFromtheGroundUp May 29 '15 at 21:32
  • No, I didn't do that. I was trying to figure out what you're trying to understand. I'm posting an answer now that will hopefully help. – Brian Tung May 29 '15 at 21:40
  • @mathFromtheGroundUp If you're curious about a more geometrical definition of the cross product, you might be interested in picking up the book Linear and Geometric Algebra by Alan Macdonald. It starts at the very beginning in linear algebra and doesn't get bogged down in the lemma, lemma, theorem, repeat style that more "rigorous" books take. It just emphasizes geometry and intuitive concepts the whole way through. The only drawback to studying it on your own is that there are no solutions either at the back of the book or available in a solutions manual. –  May 30 '15 at 00:26
  • Also, if Brian answered your question you should accept it by clicking on the checkmark at the top left of his answer. This way everyone will know that this question has already been satisfactorily answered. If he didn't answer you question well enough edit your question to make clear what else you'd like clarified. –  May 30 '15 at 00:29
  • @Bye_World: Thanks for having my back. :-) – Brian Tung May 30 '15 at 01:18
  • @Bye_World, Excellent suggestion thank you. I'll add it to the pile. XD – mathFromtheGroundUp May 31 '15 at 19:59
  • Yes, to your edit, that's right. – Brian Tung May 31 '15 at 22:34

1 Answers1

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It will be easier to understand, perhaps, if we consider two vectors $\vec{a}$ and $\vec{b}$ restricted to the $x$-$y$ plane, so that $\vec{a} = a_i \hat{i} + a_j \hat{j}$, and $\vec{b} = b_i \hat{i} + b_j \hat{j}$. Then $\vec{a} \times \vec{b} = (a_i b_j - a_j b_i) \hat{k}$, and $\|\vec{a} \times \vec{b}\| = a_i b_j - a_j b_i$. A geometrical interpretation of this quantity can be seen as the difference between two rectangles:

enter image description here

(If you admit signed areas, it will still all work out.) We can then move two triangles over to the space left by the smaller rectangle:

enter image description here

And finally, we move two (differently shaped) triangles up to the previously uncovered part of the parallelogram:

enter image description here

Does that help?

mathFromtheGroundUp
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Brian Tung
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