How to prove the cross product of two vectors?

Question

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:

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

Answer 1

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 $\hat k$: $$ (\hat i\times\hat i) = (\hat j\times \hat j) = (\hat k\times \hat k) = 0 \\ (\hat i\times \hat j) = \hat k,\;(\hat j\times \hat k) = \hat i,\;(\hat k\times \hat i) = \hat j\\ (\hat j\times \hat i) = -\hat k,\;(\hat k\times \hat j) = -\hat i,\;(\hat i\times \hat k) = -\hat j\\ $$

Answer 2

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 $$|(u,v,w)|=|A||B|\sin(\alpha).$$