The vector product of two vectors

Consider two vectors a⃗\vec{a} and b⃗\vec{b}. The vector product (or cross-product) of a⃗\vec{a} and b⃗\vec{b}, written a⃗×b⃗\vec{a} \times \vec{b} is a new vector that is defined as follows

a⃗×b⃗=(aybz−azbyazbx−axbzaxby−aybx)\vec{a}\times \vec{b} = \left(\begin{array}{r} a_y b_z-a_z b_y \\ a_z b_x-a_x b_z \\ a_x b_y - a_y b_x \end{array}\right)

There are several methods for finding the components of the new vector. Here is one:

Example 1

Determine the vector product a⃗×b⃗\vec{a}\times \vec{b}:

  1. a⃗=(1−23)\vec{a}=\left(\begin{array}{r} 1 \\ -2 \\ 3 \end{array}\right), b⃗=(43−1)\vec{b}=\left(\begin{array}{r} 4 \\ 3 \\ -1 \end{array}\right)

  2. a⃗=(100)\vec{a}=\left(\begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right), b⃗=(010)\vec{b}=\left(\begin{array}{r} 0 \\ 1 \\ 0 \end{array}\right)

Solution

The most important property of the vector product is that the vector a⃗×b⃗\vec{a}\times \vec{b} is orthogonal to both vectors a⃗\vec{a} and b⃗\vec{b} (see figure below):

(a⃗×b⃗)⊥a⃗  and  (a⃗×b⃗)⊥b⃗\boxed{(\vec{a}\times \vec{b}) \perp \vec{a} \, \text{ and }\, (\vec{a}\times \vec{b}) \perp \vec{b}}

The prove is left as an exercise below.

A typical use case of the first property is to find the normal vector n⃗\vec n of a plane that is given by three points A,BA, B and CC (see figure below).

Clearly we have

n⃗=AB→×AC→\boxed{ \vec n = \overrightarrow{AB} \times \overrightarrow{AC}}
Example 2

A plane EE contains the three points A(1∣−1∣2),B(4∣2∣−1)A(1\vert -1\vert 2), B(4\vert 2\vert -1) and C(−1∣2∣4)C(-1 \vert 2\vert 4). Find the normal equation of EE.

Solution

We first find a normal vector n⃗\vec n of EE:

n⃗=AB→×AC→=(33−3)×(−232)=(15015)\vec n = \overrightarrow{AB} \times \overrightarrow{AC} = \left(\begin{array}{r} 3 \\ 3 \\ -3 \end{array}\right) \times \left(\begin{array}{r} -2 \\ 3 \\ 2 \end{array}\right) = \left(\begin{array}{r} 15 \\ 0\\ 15 \end{array}\right)

We can choose this vector, or any other that is collinear to this one, e.g.

n⃗=(101)\vec n = \left(\begin{array}{r} 1 \\ 0\\ 1 \end{array}\right)

We take the latter one, as there is less to calculate.

We need a point on EE to find the normal equation. With point AA we get d=1+0+2=3d=1+0+2=3 and therefore the normal equation is

x+z=3x+z=3

Note that we could have taken BB or CC as well, with the same result.

Another useful property concerns the magnitude of a⃗×b⃗\vec a \times \vec b:

∣a⃗×b⃗∣=∣a⃗∣⋅∣b⃗∣⋅sin⁡(γ)\boxed{\vert \vec{a}\times \vec{b}\vert =\vert \vec a\vert \cdot \vert \vec b\vert \cdot \sin(\gamma)}

Note that the right side of this equation is simply the formula for calculating the area AA of the parallelogram whose sides are a⃗\vec a and b⃗\vec b (see figure below). This proof is rather technical, and is omitted.

Example 3

Consider the points A(2∣1∣−2),B(3∣3∣1),C(−3∣2∣0)A(2\vert 1\vert -2), B(3\vert 3\vert 1), C(-3\vert 2\vert 0) and D(−2∣4∣3)D(-2\vert 4\vert 3).

  1. Determine the area of the parallelogram ABCDABCD.

  2. Determine the area of the triangle ABCABC.

Solution

  1. First we calculate the vector product

    AB→×AC→=(123)×(−512)=(1−1711)\overrightarrow{AB} \times \overrightarrow{AC} = \left(\begin{array}{r} 1 \\ 2 \\ 3 \end{array}\right) \times \left(\begin{array}{r} -5 \\ 1 \\ 2 \end{array}\right) = \left(\begin{array}{r} 1 \\ -17\\ 11 \end{array}\right)

    Thus, the area of the parallelogram is

    A=∣AB→×AC→∣=12+(−17)2+112=20.273‾A=\vert \overrightarrow{AB} \times \overrightarrow{AC}\vert = \sqrt{1^2+(-17)^2+11^2}=\underline{20.273}

  2. The triangle area is just halve the area of the parallelogram (see figure below), so we have

    A=12∣AB→×AC→∣=10.137‾A=\frac{1}{2}\vert \overrightarrow{AB} \times \overrightarrow{AC}\vert = \underline{10.137}
Exercise 1
Q1

Proof that (a⃗×b⃗)⊥a⃗(\vec a \times \vec b) \perp \vec a and (a⃗×b⃗)⊥b⃗(\vec a \times \vec b) \perp \vec b

Q2

A plane EE contains the three points A(1∣−1∣2)A(1\vert -1 \vert 2), B(−1∣1∣1)B(-1\vert 1\vert 1), and C(4∣−2∣5)C(4\vert -2\vert 5). Determine

  1. the normal equation of EE.

  2. the area of the triangle ABCABC.

Solution
A1

We have to show that (a⃗×b⃗)∙a⃗=0(\vec{a}\times \vec{b})\bullet \vec{a}=0.

(a⃗×b⃗)∙a⃗=(aybz−azbyazbx−axbzaxby−aybx)∙(axayaz)=aybzax−azbyax+azbxay−axbzay+axbyaz−aybxaz=0\begin{array}{lll} (\vec{a}\times \vec{b})\bullet \vec a & = & \left(\begin{array}{r} a_y b_z-a_z b_y \\ a_z b_x-a_x b_z \\a_x b_y - a_y b_x\end{array}\right) \bullet \left(\begin{array}{r} a_x \\ a_y\\ a_z \end{array}\right)\\ & = & a_y b_z a_x -a_z b_y a_x + a_z b_x a_y -a_x b_z a_y + a_x b_y a_z - a_y b_x a_z \\ & = & 0\end{array}

A similar calculation shows that (a⃗×b⃗)∙b⃗=0(\vec{a}\times \vec{b})\bullet \vec{b}=0.

A2

  1. A normal vector of EE is

    n⃗=AB→×AC→=(−22−1)×(3−13)=(53−4)\vec{n}=\overrightarrow{AB} \times \overrightarrow{AC} = \left(\begin{array}{r} -2 \\ 2 \\ -1 \end{array}\right) \times \left(\begin{array}{r} 3 \\ -1 \\ 3 \end{array}\right) = \left(\begin{array}{r} 5 \\ 3\\ -4 \end{array}\right)

    With point AA in EE, we get d=−6d=-6. Thus, the normal equation is

    5x+3y−4y=−65x+3y-4y=-6

  2. We use the magnitude of the vector product:

    A=12∣AB→×AC→∣=12∣n⃗∣=1252+32+(−4)2=3.536\begin{array}{lll} A&=& \frac{1}{2}\vert \overrightarrow{AB} \times \overrightarrow{AC}\vert \\ &=& \frac{1}{2}\vert \vec n\vert \\ &=& \frac{1}{2}\sqrt{5^2+3^2+(-4)^2} \\ &=& 3.536 \end{array}