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.


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.


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.

  • $\begingroup$ How do you define the cross product? $\endgroup$
    – user137731
    Commented May 29, 2015 at 20:56
  • $\begingroup$ a = a1i + a2j + a3z. b = b1i + b2j + b3z. a x b = (a2*b3 - a3*b2)i + (a3*b1 - a1*b3)j + (a1*b2 - a2*b1)k. Where 'a' and 'b' are vectors, with the components as shown. $\endgroup$ Commented May 29, 2015 at 20:59
  • $\begingroup$ 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. $\endgroup$ Commented May 29, 2015 at 21:01
  • $\begingroup$ 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? $\endgroup$
    – user137731
    Commented May 29, 2015 at 21:02
  • $\begingroup$ Right, that's the basic definition. But not why those two are equal or how one would show it with geometry. @StevenVanGeluwe $\endgroup$ Commented May 29, 2015 at 21:02

1 Answer 1


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?

  • $\begingroup$ By the way, I apologize for the revolting color scheme. Best I can do on short notice. :-) $\endgroup$
    – Brian Tung
    Commented May 29, 2015 at 21:50
  • $\begingroup$ Wow, much better than my efforts with pen and paper. Let me ponder on that a bit, thank you very much. =D $\endgroup$ Commented May 29, 2015 at 21:53
  • $\begingroup$ That was great @BrianTung! Been looking a lot for that and only found it here :) Do you know of a similar argument for the 3x3 determinant and volumes? $\endgroup$
    – Lucas Seco
    Commented Jul 8, 2016 at 14:56
  • $\begingroup$ I guess for this solution, we need ai to be >= 2*bi, and bj to be >= 2*aj, but it's a nice one, thanks. $\endgroup$
    – aderchox
    Commented Jan 26, 2021 at 16:29

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