Geometric understanding of the Cross Product

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

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

• I'm not looking for a derivation of the formula! I'm looking for a geometric connection between determinants and the formula...
– dfg
Dec 6 '13 at 23:09
• 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. Dec 6 '13 at 23:13
• I don't see it. Could you please make it more explicit?
– dfg
Dec 6 '13 at 23:20
• 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.
– 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.

• 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.
– 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 $$A$$.

QED!