Cross product of vectors

The cross product can be used to determine the angle $\theta$ between vectors $\mathbf{a}$ and $\mathbf{b}$. Given $\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \mathbf{\hat{n}}$, where $\mathbf{\hat{n}}$ is a unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$, we can find the sine of the angle:

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

Therefore, $\theta = \sin^{-1}\left(\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}| |\mathbf{b}|}\right)$

Example 1

Find the angle between the vectors $\mathbf{a} = 4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 3\mathbf{j} - \mathbf{k}$.

Solution:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -2 & 5 \\ 1 & 3 & -1 \end{vmatrix} = ((-2)(-1) - 5(3))\mathbf{i} - (4(-1) - 5(1))\mathbf{j} + (4(3) - (-2)(1))\mathbf{k}$

$\mathbf{a} \times \mathbf{b} = (2 - 15)\mathbf{i} - (-4 - 5)\mathbf{j} + (12 + 2)\mathbf{k} = -13\mathbf{i} + 9\mathbf{j} + 14\mathbf{k}$

$|\mathbf{a} \times \mathbf{b}| = \sqrt{(-13)^2 + 9^2 + 14^2} = \sqrt{169 + 81 + 196} = \sqrt{446}$

$|\mathbf{a}| = \sqrt{4^2 + (-2)^2 + 5^2} = \sqrt{16 + 4 + 25} = \sqrt{45}$

$|\mathbf{b}| = \sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11}$

$\sin\theta = \frac{\sqrt{446}}{\sqrt{45}\sqrt{11}} = \frac{\sqrt{446}}{\sqrt{495}} \approx \sqrt{0.9010} \approx 0.9492$

$\theta = \sin^{-1}(0.9492) \approx 71.6^\circ$

Example 2

Find the angle between $\mathbf{a}$ and $\mathbf{b}$, given that $\mathbf{a} = 2\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}$ and $\mathbf{b} = 6\mathbf{i} + 8\mathbf{j}$.

Solution:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 4 & 8 \\ 6 & 8 & 0 \end{vmatrix} = (4(0) - 8(8))\mathbf{i} - (2(0) - 8(6))\mathbf{j} + (2(8) - 4(6))\mathbf{k}$

$\mathbf{a} \times \mathbf{b} = -64\mathbf{i} + 48\mathbf{j} - 8\mathbf{k}$

$|\mathbf{a} \times \mathbf{b}| = \sqrt{(-64)^2 + 48^2 + (-8)^2} = \sqrt{4096 + 2304 + 64} = \sqrt{6464} = 8\sqrt{101}$

$|\mathbf{a}| = \sqrt{2^2 + 4^2 + 8^2} = \sqrt{4 + 16 + 64} = \sqrt{84} = 2\sqrt{21}$

$|\mathbf{b}| = \sqrt{6^2 + 8^2 + 0^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

$\sin\theta = \frac{8\sqrt{101}}{10 \cdot 2\sqrt{21}} = \frac{4\sqrt{101}}{10\sqrt{21}} = \frac{2\sqrt{101}}{5\sqrt{21}} \approx \frac{20.1}{22.9} \approx 0.8777$

$\theta = \sin^{-1}(0.8777) \approx 61.4^\circ$

The cross product of vectors can be used to find the areas of a triangle and a parallelogram.

The area of a parallelogram formed by vectors $\mathbf{a}$ and $\mathbf{b}$ is given by:

$A = |\mathbf{a} \times \mathbf{b}|$

Since $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\theta$, where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$, the area can also be written as:

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

The area of a triangle formed by vectors $\mathbf{a}$ and $\mathbf{b}$ is half the area of the parallelogram formed by the same vectors:

$A = \frac{1}{2}|\mathbf{a} \times \mathbf{b}|$

Area of a parallelogram using diagonal vectors

Consider a parallelogram with adjacent sides formed by vectors $\mathbf{a}$ and $\mathbf{b}$, and diagonals $\mathbf{d}_1$ and $\mathbf{d}_2$ (Insert Figure 2.14 here). We have:

$\mathbf{d}_1 = \mathbf{a} + \mathbf{b}$

$\mathbf{d}_2 = \mathbf{b} - \mathbf{a}$

Taking the cross product of the diagonals:

$\mathbf{d}_1 \times \mathbf{d}_2 = (\mathbf{a} + \mathbf{b}) \times (\mathbf{b} - \mathbf{a}) = \mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{a} + \mathbf{b} \times \mathbf{b} - \mathbf{b} \times \mathbf{a}$

Since $\mathbf{a} \times \mathbf{a} = \mathbf{0}$ and $\mathbf{b} \times \mathbf{b} = \mathbf{0}$, and $\mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b}$:

$\mathbf{d}_1 \times \mathbf{d}_2 = \mathbf{a} \times \mathbf{b} - (-\mathbf{a} \times \mathbf{b}) = 2(\mathbf{a} \times \mathbf{b})$

Therefore, the area of the parallelogram in terms of its diagonals is:

$A = \frac{1}{2}|\mathbf{d}_1 \times \mathbf{d}_2|$

Example 1

Find the area of a triangle formed by the points $P_1(2, 3, 4)$, $P_2(1, 1, 1)$, and $P_3(3, 0, -1)$.

Solution:

$\overrightarrow{P_1P_2} = (1-2)\mathbf{i} + (1-3)\mathbf{j} + (1-4)\mathbf{k} = -\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}$

$\overrightarrow{P_1P_3} = (3-2)\mathbf{i} + (0-3)\mathbf{j} + (-1-4)\mathbf{k} = \mathbf{i} - 3\mathbf{j} - 5\mathbf{k}$

$\overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -2 & -3 \\ 1 & -3 & -5 \end{vmatrix} = (10 - 9)\mathbf{i} - (5 - (-3))\mathbf{j} + (3 - (-2))\mathbf{k} = \mathbf{i} - 8\mathbf{j} + 5\mathbf{k}$

$|\overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}| = \sqrt{1^2 + (-8)^2 + 5^2} = \sqrt{1 + 64 + 25} = \sqrt{90} = 3\sqrt{10}$

Area of the triangle: $A = \frac{1}{2} \cdot 3\sqrt{10} = \frac{3\sqrt{10}}{2}$ square units.

Example 2

The adjacent sides of a triangle are formed by the vectors $\mathbf{a} = 4\mathbf{i} - 5\mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}$. Find the area of the triangle.

Solution:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -5 & 3 \\ 2 & 2 & -4 \end{vmatrix} = (20-6)\mathbf{i} - (-16-6)\mathbf{j} + (8-(-10))\mathbf{k} = 14\mathbf{i} + 22\mathbf{j} + 18\mathbf{k}$

$|\mathbf{a} \times \mathbf{b}| = \sqrt{14^2 + 22^2 + 18^2} = \sqrt{196 + 484 + 324} = \sqrt{1004} = 2\sqrt{251}$

Area of the triangle: $A = \frac{1}{2} \cdot 2\sqrt{251} = \sqrt{251}$ square units.

Example 3

The adjacent sides of a parallelogram are defined by the vectors $\mathbf{a} = 8\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = -2\mathbf{j} + \mathbf{k}$. Find the area of the parallelogram.

Solution:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 8 & 2 & -1 \\ 0 & -2 & 1 \end{vmatrix} = (2 - 2)\mathbf{i} - (8 - 0)\mathbf{j} + (-16 - 0)\mathbf{k} = 0\mathbf{i} - 8\mathbf{j} - 16\mathbf{k} = -8\mathbf{j}-16\mathbf{k}$

$|\mathbf{a} \times \mathbf{b}| = \sqrt{0^2 + (-8)^2 + (-16)^2} = \sqrt{0 + 64 + 256} = \sqrt{320} = 8\sqrt{5}$

Area of the parallelogram: $A = 8\sqrt{5}$ square units.

Let $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ be vectors representing the sides of a triangle ABC, as shown in Figure 2.15 (Insert Figure 2.15 here). The sides of the triangle are related by the vector equation:

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

(or equivalently, $\mathbf{c} = -(\mathbf{a} + \mathbf{b})$ or $\mathbf{a} = -(\mathbf{b} + \mathbf{c})$ or $\mathbf{b} = -(\mathbf{a}+\mathbf{c})$)

Which is equivalent to $\mathbf{a} + \mathbf{b} = -\mathbf{c}$

Taking the cross product of both sides of $\mathbf{a} + \mathbf{b} = -\mathbf{c}$ with $\mathbf{a}$:

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

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

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

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

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

Similarly, taking the cross product of $\mathbf{a} + \mathbf{b} = -\mathbf{c}$ with $\mathbf{b}$:

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

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

Since $\mathbf{b} \times \mathbf{b} = \mathbf{0}$:

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

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

Using the property that $\mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b}$:

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

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

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

Combining the results, we have:

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

Taking the magnitudes:

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

Using the definition of the magnitude of the cross product: $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}| |\mathbf{v}| \sin\theta$, where $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$:

$|\mathbf{a}| |\mathbf{b}| \sin C = |\mathbf{b}| |\mathbf{c}| \sin A = |\mathbf{c}| |\mathbf{a}| \sin B$

Dividing by $|\mathbf{a}| |\mathbf{b}| |\mathbf{c}|$:

$\frac{\sin C}{|\mathbf{c}|} = \frac{\sin A}{|\mathbf{a}|} = \frac{\sin B}{|\mathbf{b}|}$

Which is commonly written as:

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

Where $a$, $b$, and $c$ are the magnitudes of the vectors (side lengths of the triangle).