Why does the determinate definition of the cross product give a vector that is perpendicular to the plane?

Question

I'm trying to understand what a cross product really is. From what I can tell, the length of $\mathbf a \times \mathbf b$ is a measurement of how much $\mathbf a$ and $\mathbf b$ are NOT moving together. But that doesn't really help me understand what $\mathbf a \times \mathbf b$ is. Why does the cross product give a perpendicular vector?

Answer 1

HINT: Compute scalar product of $\vec{a}\times\vec{b}$ and $\vec{a}$. (I am assuming that you have in mind the standard 3-dimensional case).

Answer 2

The "determinate (?) definition of the cross product", like so: $${\bf a}\times{\bf b}:=\det\left[\matrix{{\bf i}&{\bf j}&{\bf k}\cr a_1&a_2&a_3\cr b_1&b_2&b_3\cr}\right]\ ,$$ is crap for dummies. In the first place the cross product ${\bf a}\times{\bf b}$ of two vectors ${\bf a}$, ${\bf b}$ in euclidean three space is a vector encoding certain geometric information about the pair $({\bf a},{\bf b})$ which is associated to this pair in a geometrically invariant way, namely (a) the normal to the (oriented!) plane which these vectors span, and (b) the area of the parallelogram which they span. That this encoding can be done in a bilinear way in terms of the data ${\bf a}$, ${\bf b}$ is a geometric miracle which should be accepted with grace and not considered as a nuisance. It is only around 1800 AC, after 2500 years of spatial geometry, that mathematicians and physicists finally nailed down this concept, see the article Cross Product in Wikipedia.

Answer 3

The determinant 'definition' doesn't need a lot to show that the result is perpendicular to the two vectors you cross together. Let $\vec{v}=v_x\vec{i}+v_y\vec{j}+v_z\vec{k}$ and similar for $\vec{w}$

$\begin{vmatrix} x & y & z \\ v_i & v_j & v_k \\ w_i & w_j & w_k \end{vmatrix} = \begin{vmatrix} v_j & v_k \\ w_j & w_k \end{vmatrix}x - \begin{vmatrix} v_i & v_k \\ w_i & w_k \end{vmatrix}y +\begin{vmatrix} v_i & v_j \\ w_i & w_j \end{vmatrix}z = \left( \begin{vmatrix} v_j & v_k \\ w_j & w_k \end{vmatrix}, -\begin{vmatrix} v_i & v_k \\ w_i & w_k \end{vmatrix}, \begin{vmatrix} v_i & v_j \\ w_i & w_j \end{vmatrix}\right)\cdot\left(x,y,z\right) $.

Now recall that if a matrix has two rows the same, then it has determinant zero. The above then means that this dot product will be zero if x,y,z are the components of either v or w, so the vector whose components are the 2x2 determinants will be perpendicular to both v and w.