Question

is there a way to show this identity other than using brute force?

\[\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})=\left|\begin{matrix}A_x&A_y&A_z\\B_x&B_y&B_z\\C_x&C_y&C_z\end{matrix}\right|\]

Answer 1

yes. This represents a volume of a paralelepipedon.

http://www.vitutor.com/geometry/vectors/tiple_product.html

Answer 2

I know that. but how do I show that this equality holds without writing out that cross product and then dotting it with A. I am trying to show that this equality is true.

Answer 3

I don't think there is a way. Just imagine geometrical picture. Projection of BxC on A, and this is enough, at least for me, to remmember it,:). Maybe there is other way, but i dont know.

Answer 4

thanks. I do not need to remember this though. I learned this in first or second year math. my question specifically asks me to show it so I better figure it out!

Answer 5

then just use definition

Answer 6

\(B \times C= (B_yC_z-B_zC_y, B_zC_x-B_xC_z,B_xC_y-B_yC_x)\)
so:
\(A \cdot (B \times C)=A_x(B_yC_z-B_zC_y)+A_y( B_zC_x-B_xC_z)+A_z(B_xC_y-B_yC_x)\)
and this expantion by elements of 1º row of the determinant that you have

Answer 7

@richyw