Introduction to 3D Cross Product

The 3D cross product (aka 3D outer product or vector product) of two vectors $\mathbf{a}$ and $\mathbf{b}$ is only defined on three dimensional vectors as another vector $\mathbf{a}\times\mathbf{b}$ that is orthogonal to the plane containing both $\mathbf{a}$ and $\mathbf{b}$ and has a magnitude of

$$|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}| |\mathbf{b}| \sin\theta$$

where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$. The length of the cross product equals the area of the parallelogram bordered by the two vectors. Also note that $\sin\theta$ is always positive, since the angle will never exceed 180°.

Note that if $\mathbf{a}$ and $\mathbf{b}$ have a length of one, the length of the cross product is only one when $\theta=90^\circ$, as $\sin\theta=1$.

Another definition of the cross-product keeps the direction of the cross product in a unit vector $\hat{\mathbf{n}}$ perpendicular to $\mathbf{a}$ and $\mathbf{b}$ such that:

$$\mathbf{a}\times\mathbf{b}=\hat{\mathbf{n}}|\mathbf{a}||\mathbf{b}|\sin\theta$$

From which is also obvious that collinear (parallel) vectors $\mathbf{a}$ and $\mathbf{b}$ result in $\sin\theta=0$ and thus $\mathbf{a}\times\mathbf{b} = \mathbf{0}$

To derive the cross product $\mathbf{c} = \mathbf{a}\times\mathbf{b}$. From the definition that $\mathbf{c}$ is orthogonal to both, $\mathbf{a}$ and $\mathbf{b}$, we know that

$$\mathbf{a}\cdot\mathbf{c} = a_xc_x+a_yc_y+a_zc_z=0$$

$$\mathbf{b}\cdot\mathbf{c} = b_xc_x+b_yc_y+b_zc_z=0$$

To eliminate $c_z$ we multiply the first equation by $b_z$ and the second by $a_z$ and subtract:

$$(a_xb_z-a_zb_x)c_x + (a_yb_z-a_zb_y)c_y=0$$

This equation has the form $pc_x+qc_y=0$, for which an obvious solution is $c_x=q$ and $c_y=-p$. Thus

$$c_x=a_yb_z-a_zb_y$$$$c_y=a_zb_x-a_xb_z$$

and by substituting in the original equations:

$$c_z=a_xb_y-a_yb_x$$

Alternatively, we can derive the cross product using the determinant, similar to the calculation of the 2D perp-product or perp-roduct using the determinant with $\hat{\mathbf{x}}, \hat{\mathbf{y}}$ and $\hat{\mathbf{z}}$ being the orthonormal basis.:

$$\begin{array}{rl} \mathbf{a}\times\mathbf{b} &= \left|\begin{array}{ccc}\hat{\mathbf{x}}&\hat{\mathbf{y}}&\hat{\mathbf{z}}\\ a_x & a_y & a_z\\ b_x & b_y & b_z \end{array} \right|\\ &= \hat{\mathbf{x}}\left|\begin{array}{cc} a_y & a_z\\ b_y & b_z \end{array} \right| - \hat{\mathbf{y}}\left|\begin{array}{cc} a_x & a_z\\ b_x & b_z \end{array} \right| + \hat{\mathbf{z}}\left|\begin{array}{cc} a_x & a_y\\ b_x & b_y \end{array} \right|\\ &= \hat{\mathbf{x}}(a_yb_z-a_zb_y) - \hat{\mathbf{y}}(a_zb_x-a_xb_z) + \hat{\mathbf{z}}(a_xb_y-a_yb_x)\\ &= \left(\begin{array}{c} a_yb_z-a_zb_y\\ a_zb_x-a_xb_z\\ a_xb_y-a_yb_x \end{array}\right) \end{array}$$

Cross Product Properties

There are two possible choices to compute the cross product, each the negation of the other. This makes the cross product not commutative and thus anticommutative / antisymmetric. The one chosen is determined by the right-hand rule. If your index finger is $\mathbf{a}$, your middle finger $\mathbf{b}$ then your thumb is the positive cross product $\mathbf{a}\times\mathbf{b}$.

$$\mathbf{a}\times\mathbf{b}=-(\mathbf{b}\times\mathbf{a}) = (-\mathbf{b})\times\mathbf{a}$$

Additive Distribution

$$\mathbf{a}\times(\mathbf{b}+\mathbf{c}) = \mathbf{a}\times\mathbf{b}+\mathbf{a}\times\mathbf{c}$$

$$(\mathbf{a}+\mathbf{b})\times\mathbf{c} = \mathbf{a}\times\mathbf{c}+\mathbf{b}\times\mathbf{c}$$

Großmann Identity or Double vector product

Left Association

$$(\mathbf{a}\times\mathbf{b})\times\mathbf{c} = (\mathbf{a}\cdot\mathbf{c})\mathbf{b} - (\mathbf{b}\cdot\mathbf{c})\mathbf{a}$$

Right Association

$$\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\cdot\mathbf{c})\mathbf{b} - (\mathbf{a}\cdot\mathbf{b})\mathbf{c}$$

Lie Identity

$$\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) + \mathbf{c}\times(\mathbf{a}\times\mathbf{b}) + \mathbf{b}\times(\mathbf{c}\times\mathbf{a}) = \mathbf{0}$$

Dot-Cross Association

$$\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}$$

Scalar Association

$$(\alpha\mathbf{a})\cdot(\beta\mathbf{b}) = (\alpha\beta)(\mathbf{a}\times\mathbf{b})$$

$$\alpha(\mathbf{a}\times\mathbf{b}) = (\alpha\mathbf{a})\times\mathbf{b} = \mathbf{a}\times(\alpha\mathbf{b})$$

Normality

$$(\mathbf{a}\times\mathbf{b})\cdot\mathbf{a} = (\mathbf{a}\times\mathbf{b})\cdot\mathbf{b} = 0$$

Nilpotent

$$\mathbf{a}\times\mathbf{a}=\mathbf{0}$$

$$\mathbf{a}\times\mathbf{0} = \mathbf{0}\times\mathbf{a} = \mathbf{0}$$

Jacobi Identity

$$\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) + \mathbf{b}\times(\mathbf{c}\times\mathbf{a}) + \mathbf{c}\times(\mathbf{a}\times\mathbf{b}) = \mathbf{0}$$

Cyclic Permutations

The following cyclic permutations of cross products for the vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ hold:

$$\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}$$

$$\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a}=\mathbf{0}$$

Lagrange Identity

$$(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{c}\times\mathbf{d}) = (\mathbf{a}\cdot\mathbf{c})(\mathbf{b}\cdot\mathbf{d}) - (\mathbf{a}\cdot\mathbf{d})(\mathbf{b}\cdot\mathbf{c})$$

From which follows that the square of the norm is

$$|\mathbf{a}\times\mathbf{b}|^2 = |\mathbf{a}|^2|\mathbf{b}|^2 - (\mathbf{a}\cdot\mathbf{b})^2$$

which is a nice connection between the dot-product and the cross product. Especially with normalized vectors this is

$$\underbrace{|\hat{\mathbf{a}}\times\hat{\mathbf{b}}|^2}_{=\sin^2\theta} = 1 - \underbrace{(\hat{\mathbf{a}}\cdot\hat{\mathbf{b}})^2}_{=\cos^2\theta}$$

But we can go further with the square of the norm:

$$\begin{array}{rl} |\mathbf{a}\times\mathbf{b}|^2 &= (\mathbf{a}\times\mathbf{b})\cdot(\mathbf{a}\times\mathbf{b})\\ &= (\mathbf{a}\cdot\mathbf{a})(\mathbf{b}\cdot\mathbf{b}) - (\mathbf{a}\cdot\mathbf{b})^2\\ &= |\mathbf{a}|^2|\mathbf{b}|^2(1-\cos^2\theta)\\ &= |\mathbf{a}|^2|\mathbf{b}|^2\sin^2\theta \end{array}$$

From which follows the definition of the cross product:

$$|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta$$

Which works since the angle between $\mathbf{a}$ and $\mathbf{b}$ is always between 0° and 180° and therefore $\sin\theta\geq 0$.

Scalar Triple Product

The dot-cross product $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$ has the name scalar triple product and is denoted by $[\mathbf{a},\mathbf{b},\mathbf{c}]$. It computes the (signed) volume of the parallelepiped spanned by the vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$:

$$\begin{array}{rl} [\mathbf{a},\mathbf{b},\mathbf{c}] &= \det(\mathbf{a}\;\mathbf{b}\;\mathbf{c})\\ &= \mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\\ &= \mathbf{b}\cdot (\mathbf{c}\times \mathbf{a})\\ &= \mathbf{c}\cdot (\mathbf{a}\times \mathbf{b})\\ \end{array}$$

The sign is positive if $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ form a right handed set and negative if they form a left handed set.

Vector Triple Product

The vector triple product is

$$\mathbf{a}\times(\mathbf{b}\times\mathbf{c})$$

Applications

Normal to a triangle

The most common application of the cross product is to generate a vector orthogonal to two other vectors. Suppose we have three points $P$, $Q$ and $R$ and want to generate a unit vector $\hat{\mathbf{n}}$ that is orthogonal to the plane formed by the three points.

Now $\mathbf{a}=Q-P$ and $\mathbf{b}=R-P$ and the normal can be found with $\mathbf{n}=\mathbf{a}\times\mathbf{b}$. The direction of the normal is usually chosen to point from the inside to the outside of our object.

Interestingly, the length $|\mathbf{n}|$ here equals twice the area of the triangle (since halve of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ is our triangle).

Equation of a plane given three points

Given three points $P_1$, $P_2$, $P_3$, we can form two vectors $\mathbf{v}_1=\vec{P_1P_2}$ and $\mathbf{v}_2=\vec{P_1P_3}$ that lie in that plane. Now let $P=(x,y,z)^T$ be a general point in space. $P$ lies in the plane iff the vector $\mathbf{v}_3=\vec{P_1P}$ lies in that plane, which is the case if and only if the volume of the parallelepiped spanned by $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ is zero and therefore the Scalar Triple Product:

$$[\mathbf{v_1},\mathbf{v_2},\mathbf{v_3}] = 0$$

This is enough to solve for the plane equation. Another method fixes a point, i.e. $P_1$, in the plane and requiring that every vector in the plane will tail $P_1$ and is normal to $\mathbf{n}$. Therefore

$$\mathbf{n} = \mathbf{v}_1\times\mathbf{v}_2$$

is one normal to the plane. Now $P$ is in the plane if and only if $\mathbf{v}_3$ is orthogonal to $\mathbf{n}$.

$$\mathbf{v}_3\cdot\mathbf{n} = 0$$

From there the typical plane equation can be read off, which has the form

$$(x-P_{1,x})\mathbf{n}_x+(y-P_{1,y})\mathbf{n}_y+(z-P_{1,z})\mathbf{n}_z=0$$

Distance between two lines

Consider two lines in space $\ell_1$ and $\ell_2$ such that point $\ell_1$ passes through $P_1$ and is parallel to vector $\mathbf{v}_1$ and $\ell_2$ passes through $P_2$ and is parallel to $\mathbf{v}_2$. We want to compute the smallest distance $d$ between the two lines.

If the lines intersect, the distance is $d=0$.

If they are parallel, then $d$ corresponds to the distance between point $P_2$ and $\ell_1$:

$$d=\frac{\|\overrightarrow{P_1P_2}\times\mathbf{v}_1\|}{\|\mathbf{v}_1\|}$$

If the lines are not paralll and do not intersect (skew lines) then let $\mathbf{n}=\mathbf{v}_1\times\mathbf{v}_2$ be a vector perpendicular to both lines. The projection of vector onto $\mathbf{n}$ gives $d$:

$$d=\frac{|\overrightarrow{P_1P_2}\cdot\mathbf{n}|}{\|\mathbf{n}\|}$$

Test if two vectors are parallel

Like the dot product, the cross product can be used to determine if two vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel, which is the case when $\mathbf{a}\times\mathbf{b}=\mathbf{0}$. This result comes directly from the definition of the length of the cross product, since $\sin(\theta)=0$ for 0° and 180°. Calculating $|\mathbf{a}\times\mathbf{b}|=0$ or simply $(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{a}\times\mathbf{b})=0$ has the advantage of using less operations with nonnormalized vectors over the parallel-test using the dot-product.