Okay, I must admit that I am lost on how to do this. I have looked up videos and tutorials about this, and they helped a little. The main thing is that my professor asked for us to solve this without using the "determinant method." I have just started linear algebra, so I am still trying to understand determinants and the like. I am just wondering how it is possible to prove the cross product of two vectors with another method? Any help would be great!

Prove the following without using the determinant method:

$A \times B = (A_yB_z - B_yA_z)\hat{i} - (A_xB_z - B_xA_z)\hat{j} + (A_xB_y - B_xA_y)\hat{k}$

  • $\begingroup$ I've editted the question for math formatting. Feel free to let me know if it doesn't reflect your intended meaning. You can find a LaTeX tutorial on this site if you need further help. $\endgroup$ – Alfred Yerger Feb 4 '17 at 4:24
  • $\begingroup$ determinant method?? $\endgroup$ – mle Feb 4 '17 at 4:28
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    $\begingroup$ Language quibble: one does not "prove the cross product of two vectors" any more than one proves apples or chairs or bicycles. You want to prove an identity about cross products. Anyway, what exactly is your definition of the cross product? @mle This. $\endgroup$ – arctic tern Feb 4 '17 at 4:29
  • $\begingroup$ from Wiki: ^ Here, “formal" means that this notation has the form of a determinant, but does not strictly adhere to the definition; it is a mnemonic used to remember the expansion of the cross product. $\endgroup$ – mle Feb 4 '17 at 4:38
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    $\begingroup$ @Luke: mle asked what you meant by "determinant method" and I responded by telling mle what you were referring to. It is I who is puzzled by mle repeating back the words from wikipedia that I just linked him or her too. Perhaps mle is suggesting that because it is not technically a determinant it should not be called a "determinant method," but this seems inconsistent with expressing confusion over what the phrase means in the first place. $\endgroup$ – arctic tern Feb 4 '17 at 4:47

The proof can be given using the distributive property of the cross product and the fact that $c(v\times w) = (cv)\times w = v\times (cw)$ for vectors $v$ and $w$ and a scalar $c$: $$ A\times B = (A_x \hat i + A_y \hat j + A_z \hat k)\times (B_x \hat i + B_y \hat j + B_z \hat k)\\ = A_xB_x(\hat i\times\hat i)+A_xB_y(\hat i\times \hat j)+A_xB_z(\hat i\times \hat k)\\ + A_yB_x(\hat j\times\hat i)+A_yB_y(\hat j\times \hat j)+A_yB_z(\hat j\times \hat k)\\ + A_zB_x(\hat k\times\hat i)+A_zB_y(\hat k\times \hat j)+A_zB_z(\hat k\times \hat k)\\ $$ Now we only need to use the obvious properties of $\hat i$,$\hat j$ and $\hat k$: $$ (\hat i\times\hat i) = (\hat j\times \hat j) = (\hat k\times \hat k) = 0 \\ (\hat i\times \hat j) = \hat k,\;(\hat j\times \hat k) = \hat i,\;(\hat k\times \hat i) = \hat j\\ (\hat j\times \hat i) = -\hat k,\;(\hat k\times \hat j) = -\hat i,\;(\hat i\times \hat k) = -\hat j\\ $$

  • $\begingroup$ This is exactly what I was looking for, that helps me understand a lot. Thanks for being competent. $\endgroup$ – Luke Feb 5 '17 at 2:13

By defiition, the cross product of $A$ and $B$ is a vector $(u,v,w)\in\mathbb{R}^3$ that is perpendicular to both of them.

So, both doth products should be zero: $$(A_x,A_y,A_z)\cdot(u,v,w)=A_xu+A_yv+A_zw=0,$$ $$(B_x,B_y,B_z)\cdot(u,v,w)=B_xu+B_yv+B_zw=0.$$

From here, you can find expressions of two of the components (say, for instance, $v$ and $w$), that depend on $A$, $B$ and the other component ($u$).

Do you know anything about the angle between $A$ and $B$, (let's call it $\alpha$)? To solve for the third component, you can use that the length of the vector you are looking for is $$|(u,v,w)|=|A||B|\sin(\alpha).$$

  • $\begingroup$ Thank you so much for your help. And no, we weren't given any information about the lengths or angles. Just need to "prove" it somehow, without using determinants. $\endgroup$ – Luke Feb 4 '17 at 5:13

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