@nincompooop

Let $\vec{v}=(v_1,v_2,v_3)$ and $\vec{w}=(w_1,w_2,w_3)$. We want to find $\vec{v}\times\vec{w}$, show it's perpendicular, and show that $\vec{v}\times\vec{w}=-\vec{w}\times\vec{v}$.

Question

@nincompooop 

Let $\vec{v}=(v_1,v_2,v_3)$ and $\vec{w}=(w_1,w_2,w_3)$. We want to find $\vec{v}\times\vec{w}$, show it's perpendicular, and show that $\vec{v}\times\vec{w}=-\vec{w}\times\vec{v}$.

kinggeorge · Answer

In matrix form, the cross product is$$\left|\begin{matrix}
\vec{i}&\vec{j}&\vec{k}\
v_1&v_2&v_3\
w_1&w_2&w_3\
\end{matrix}\right|=(v_2w_3-v_3w_2)\vec{i}-(v_1w_3-w_1v_3)\vec{j}+(v_1w_2-w_1v_2)\vec{k}$$Taking dot products with $\vec{v}$ and $\vec{w}$, we get that the cross product is perpendicular.

kinggeorge · Answer

As for the second thing we want to prove, recall that when taking the determinant, elementary row operations can change it in a subtle way. In particular, swapping two rows, multiplies the determinant by $-1$. So if we swap the bottom two rows, we get that$$\left|\begin{matrix}
\vec{i}&\vec{j}&\vec{k}\
v_1&v_2&v_3\
w_1&w_2&w_3\
\end{matrix}\right|=-\left|\begin{matrix}
\vec{i}&\vec{j}&\vec{k}\
w_1&w_2&w_3\
v_1&v_2&v_3\
\end{matrix}\right|=-\vec{w}\times\vec{v}$$

kinggeorge · Answer

I'm actually TA'ing a linear algebra class this semester, so if you feel like shooting me more related questions I'll do my best to answer them.

nincompoop · Answer

yeah, I will. thank you! 
I need some math fundamentals for my electromagnetism in physics class

kinggeorge · Answer

Well, it's bedtime for me. I'm making an effort to be a little more present on this website though, and if you message me, it should get forwarded to my email. So if I'm not on when you have a question, and you for some reason desperately want me to answer it, you can try to message me.