Say you have vectors $v$ and $w$. Let there cross product be denoted by $x$ so that:

$$v \times w = x$$

According to Wikipedia:

$$x_x = v_yw_z - v_zw_y$$ $$x_y = v_zw_x - v_xw_z$$ $$x_z = v_xw_y - v_yw_x$$

This is equivalent to saying:

$$x_x = \left|\begin{matrix}v_y&v_z\\w_y&w_z\end{matrix}\right|$$ $$x_y = \left|\begin{matrix}v_z&v_x\\w_z&w_x\end{matrix}\right|$$ $$x_z = \left|\begin{matrix}v_x&v_y\\w_x&w_y\end{matrix}\right|$$

The determinant can be interpreted as the area spanned by the column vectors. Could you give me a geometric explanation of why the area of those above vectors give the coordinates of the cross product? Thanks!

EDIT: An interpretation for a 2x2 matrix is fine too

  • $\begingroup$ An interpretation for a $2x2$ matrix is fine too. $\endgroup$
    – dfg
    Dec 7 '13 at 0:47
  • $\begingroup$ @andraiamatrix this is not what dfg says. $\endgroup$
    – Jean Marie
    Jul 1 '16 at 21:48
  • $\begingroup$ In my opinion, one of the best way to understand it is by refering to a volume interpretation using "triple product" $\endgroup$
    – Jean Marie
    Jul 1 '16 at 21:53

Let $P$ be the paralellogram spanned by the vectors $a$ and $b$ and let $\phi$ be the angle between $a$ and $b$. Then we have \begin{eqnarray*} (Area(P))^2 & = & |a|^2|b|^2(\sin(\phi))^2 \\ & = & |a|^2|b|^2-|a|^2|b|^2(\cos(\phi))^2 \\ & = &|a|^2|b|^2-(a\cdot b)^2 \\ & = & (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2\\ & = &(a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^.2 \end{eqnarray*} So the area of $P$ is equal to the length of the vector $(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-a_2b_1) = a \times b$. With the coordinates you gave.

  • $\begingroup$ I'm not looking for a derivation of the formula! I'm looking for a geometric connection between determinants and the formula... $\endgroup$
    – dfg
    Dec 6 '13 at 23:09
  • $\begingroup$ That you cn see from this derivtion. Look at the paralellogram with the angle and the vectors, and look at how to calculate the area. You will see the simularity. $\endgroup$
    – user112167
    Dec 6 '13 at 23:13
  • $\begingroup$ I don't see it. Could you please make it more explicit? $\endgroup$
    – dfg
    Dec 6 '13 at 23:20
  • $\begingroup$ I don't wanna know why the cross product gives area. I wanna know why the determinants of the smaller vectors give the coordinates of the product. $\endgroup$
    – dfg
    Dec 6 '13 at 23:21

There is a geometric interpretation of this in terms of the areas of the projections of the parallelogram $P$ spanned by $\bf{v}$ and $\bf{w}$ to the coordinate planes. Let $\bf{n}$ be a unit vector perpendicular to the plane spanned by $\bf{v}$ and $\bf{w}$. Projecting this plane to the $yz$-plane is a map which decreases area by a factor of $\bf{i}\cdot \bf{n}=\bf{n}_x$, where $\bf{i}=(1,0,0)$. The two planes intersect in a line, for which projection preserves length, and for a line orthogonal to this line of intersection, the projection multiplies length by $\bf{n}_x$, so area changes by a factor of $\bf{n}_x$. As you note, each coordinate of the cross product is the (signed) area of the parallelogram spanned by the projections of $\bf{v}$ and $\bf{w}$ to each coordinate plane (which are the column vectors you refer to). So the length of the cross product is $|\bf{n}|\cdot Area(P) = Area(P)$. So this derives this formula (modulo signs) from the property that the cross product $\bf{v}\times \bf{w}$ is perpendicular to $\bf{v}$ and $\bf{w}$, and has magnitude $|\bf{v}\times \bf{w}|=Area(P)$, up to choices of signs. The sign choices are forced by the choice of dot product, which by changing the coefficients could be of different signatures.

  • 1
    $\begingroup$ Any chance of a picture? This looks like it contains valuable information/insight, but it is densely packed. I can't get from start to finish without losing traction. $\endgroup$
    – P i
    Jun 30 '16 at 16:17

Basically we want a coordinate-free proof that if $\mathbf{n}$ is a unit normal vector of an oriented plane $\Pi$ then $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{n}$ is the signed area of the parallelogram spanned by the orthogonal projections of the two vectors $\mathbf{a}$ and $\mathbf{b}$ onto $\Pi$. (My definition of $\mathbf{a}\times\mathbf{b}$ is the unique vector whose size is the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ and whose direction is perpendicular to it, oriented according the right-hand rule.) Then we can apply this with $\mathbf{n}=\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$.

Decompose $\mathbf{a}=\mathbf{a}_{\|}+\mathbf{a}_\perp$ and $\mathbf{b}=\mathbf{b}_{\|}+\mathbf{b}_\perp$ as parallel and perpendicular components with respect to the plane $\Pi$. Then FOIL out the terms of $\mathbf{a}\times\mathbf{b}$ and notice that $\mathbf{a}_{\|}\times \mathbf{b}_\perp$ and $\mathbf{a}_\perp\times\mathbf{b}_{\|}$ are both orthogonal to $\mathbf{n}$ (since the $\perp$ components are parallel to $\mathbf{n}$) hence vanish when the dot product with $\mathbf{n}$ is applied, and $\mathbf{a}_\perp\times\mathbf{b}_\perp=\mathbf{0}$ since the vectors are parallel, leaving at last the equality $(\mathbf{a}\times\mathbf{b})\cdot\mathbf{n}=(\mathbf{a}_{\|}\times\mathbf{b}_{\|})\cdot\mathbf{n}$. Observe $\mathbf{a}_{\|}\times\mathbf{b}_{\|}$ is parallel to $\mathbf{n}$ and its signed length is the signed area of the parallelogram of $\mathbf{a}_{\|}$ and $\mathbf{b}_{\|}$ spanned within $\Pi$, so the result follows.


There is indeed a geometric interpretation of $\bf{u}\times\bf{v}$ in terms of the areas of the projections of the parallelogram $\bf{P}$ spanned by $\bf{v}$ and $\bf{w}$ onto the coordinate planes.

I'll start from scratch.

Motivating problem: We wish to create a vector perpendicular to u,v, i.e. construct w s.t. $\bf{w}\cdot \bf{u} = \bf{w}\cdot \bf{v} = 0$.

Two equations in 3 unknowns: we can derive w as $\lambda(u_2 v_3 - u_3 v_2,$ [3&1], [1&2]$)$.

So let's define $\bf{u}\times \bf{v}$ as ($u_2 v_3 - u_3 v_2$, [3&1], [1&2]).

Observe $u_2 v_3 - u_3 v_2$ is just the signed area of the $(u_2, u_3)$, $(v_2,v_3)$ parallelogram. (2D determinant -- you can work it out with triangles). So we can rewrite our definition:

Define $\bf{u}\times \bf{v}$ as $(A_{yz}, A_{zx}, A_{xy})$ where $A_{\text{plane}}$ = signed area of u,v parallelogram projected onto that plane

Notice by symmetry, switching u and v just changes the sign of $u_p v_q - u_q v_p$, so we have:

$\bf{u}\times \bf{v} = -\bf{v}\times \bf{u}$

Notice also:

$\bf{i} \times \bf{j} = \bf{k}$

So we get our 'Right Hand Rule'.

Now it makes sense to ask: "Could we have skipped the algebra?" i.e. arrived at this definition purely from geometric insight. And the answer is yes!

Let n be the unit vector perpendicular to u,v.

WAIT! There is a problem here -- there are 2 choices: -n also is a valid candidate.

So if you are given i,j your choices are k,-k. So let's choose k. Looking at our right-hand, i for thumb, j for index finger and k corresponds to the direction the next finger is pointing. So we have an orientation. Let's call this the Right Hand Rule.

So let's choose the n that satisfies the same orientation.

We can show that the ratio of the area of the uv-parallelogram to it's yz projection gives $\bf{n}_x$, similarly for xz$\to \bf{n}_y$ and xy$\to \bf{n}_z$:

Here is a simplification that will illustrate this...

Let's say we have y as depth, z going upwards. We have some near-flat plane (so the normal is pointing nearly straight up). Suppose we draw a unit grid on it. Now we are going to drop each point onto the xy-plane so $(x,y,z)\to(x,y,0)$ and calculate the change in area. Furthermore let's say we have rotated things so that our plane's normal vector has depth component = 0.

So alternatively we could set this up by imagining two xy-planes. And rotate one of them slightly around the y-axis, give it a unit grid, and "drop" this grid onto the other (axis-aligned) plane.

We want the area of a projected grid square.

It should be obvious that the depth-component (y) of any point on our plane is unchanged by this projection. And a simple calculation reveals that the x-component is just $cos(\theta)$, where $\theta$ is the angle between the normals, otherwise known as $\bf{k}\cdot \bf{n}=\bf{n}_z$, where $\bf{k}=(0,0,1)$.

So, the xy-area multiplier is $\bf{n}_z$ as required: $A\bf{n}_z = A_{xy}$

Let's write these parallelogram areas using $A$.

So we have: $A_{xy} = A \bf{n}_z$, sim. for y & z.

So $A\bf{n} = (A_{yz}, A_{zx}, A_{xy})$

(EDIT: Bold rendering incorrectly for $A_{xy}$ in the above line, anyone?)

So if we define $\bf{u}\times \bf{v}$ as $(A_{yz}, A_{zx}, A_{xy})$ then we have $\bf{u}\times \bf{v} = A \bf{n}$, i.e. $\bf{u}\times \bf{v}$ is perpendicular to u,v and of length $A$.



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