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

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?

• Because it's defined that way. It's not really a good definition because the cross product is only defined on a product of $n-1$ vectors in $n$-dimensional space. But it's what we've got. If you'd like a better product that does a similar job as the cross product, look up the exterior product.
– user137731
Dec 13, 2014 at 20:26
• To expand a little more on this, basically the only uniquely determined directions out of a plane (the span of two non-parallel vectors) are the normal directions (in $\Bbb R^3$). So that's why the definition is what it is. Then the right-hand rule is just an arbitrary way of picking just $1$ of those two normals.
– user137731
Dec 13, 2014 at 20:30
• But what is it about the determinate of that particular matrix that gives a perpendicular vector? Dec 13, 2014 at 20:32
• @ZacharyF The determinant definition is $\vec a \cdot (\vec b \times \vec c) = \det[\vec a\ \vec b\ \vec c]$. This implicitly defines the cross product. Basically both the determinant and the triple scalar product give you the area of a parallelopiped and the cross product is defined as the mapping between the vectors $\vec b$ and $\vec c$ which does the job.
– user137731
Dec 13, 2014 at 20:35
• I guess this just shows I haven't taken linear algebra yet haha Dec 13, 2014 at 20:37

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.

• >>"The above then means that this dot product will be zero if x,y,z are the components of either v or w". Can you elaborate on that? Dec 14, 2014 at 22:52
• Well that determinant, if $x=v_x$ (and sim. for y,z) or $x=w_x$ (again sim. for y,z), will be zero as it will have two rows the same. My calculation shows we can see that determinant as the dot product of two vectors, so if the determinant is zero, that dot product is zero. A dot product is zero iff the vectors are orthogonal, so it shows that the vector made of 2x2 determinants is orthogonal to both $v$ and $w$. Is this more clear? Dec 15, 2014 at 22:13

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).

• That is what I was thinking. I computed the dot product and got 0, which I guess is unsurprising since the cross product is perpendicular to $\mathbf a$ Dec 13, 2014 at 20:34

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.