Finding the Cross Product of 3D Vectors

Calculating It through the Magnitudes and the Angle between the Two Vectors

The process is as follows.

Let's suppose there are two vectors, \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

Let the magnitudes be \(a\) and \(b\), respectively.

Let \(\theta\) be the angle between the two vectors.

Let the unit vector which is perpendicular to the two vectors be \(\overrightarrow{c}\),

If so,

$$\overrightarrow{a} \times \overrightarrow{b} = ab\sin\theta\overrightarrow{c}$$

Finding It Using the Components

Let the vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) be defined as follows.

$$\overrightarrow{a} = (a_x\hat{i} + a_y\hat{j} + a_z\hat{k}), \overrightarrow{b} = (b_x\hat{i}+b_y\hat{j} + b_z\hat{k})$$

(\(\hat{i}\), \(\hat{j}\), \(\hat{k}\) are the unit vectors of \(x\), \(y\), \(z\) axes.)

Then, \(\overrightarrow{a} \times \overrightarrow{b}\) is given as follows.

$$\overrightarrow{a} \times \overrightarrow{b} = (a_x\hat{i} + a_y\hat{j} + a_z\hat{k}) \times (b_x\hat{i} + b_y\hat{j} + b_z\hat{k})$$

As the distributive property is satisfied, it can be expanded in this way.

$$\overrightarrow{a}\times\overrightarrow{b} = a_x b_x(\hat{i} \times \hat{i}) + a_xb_y(\hat{i} \times \hat{j}) + a_xb_z(\hat{i} \times \hat{k}) + a_yb_x(\hat{j} \times \hat{i}) + a_yb_y(\hat{j} \times \hat{j}) + a_yb_z(\hat{j} \times \hat{k}) + a_zb_x(\hat{k} \times \hat{i}) + a_zb_y(\hat{k} \times \hat{j}) + a_zb_z(\hat{k} \times \hat{k})$$

Because the cross products of the unit vectors are determined as follows,

$$\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = zero \space vector$$

$$\hat{i} \times \hat{j} = \hat{k}, \hat{j} \times \hat{i} = -\hat{k}, \hat{j} \times \hat{k} = \hat{i}, \hat{k} \times \hat{j} = -\hat{i}, \hat{k} \times \hat{i} = \hat{j}, \hat{i} \times \hat{k} = -\hat{j}$$

it can be modified as shown below.

$$\overrightarrow{a}\times\overrightarrow{b} = a_xb_y(\hat{k}) - a_xb_z(\hat{j}) - a_yb_x(\hat{k}) + a_yb_z(\hat{i}) + a_zb_x(\hat{j}) - a_zb_y(\hat{i})$$

Rearranging this, we obtain the following.

$$
\overrightarrow{a}\times\overrightarrow{b} =
(a_yb_z - a_zb_y)(\hat{i}) + (a_zb_x - a_xb_z)(\hat{j}) + (a_xb_y - a_yb_x)(\hat{k})
$$

If we look closely at the above equation, it is the determinant of the following matrix.

$$
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
a_x & a_y & a_z \\
b_x & b_y & b_z
\end{vmatrix}
=
\begin{vmatrix}
a_y & a_z \\
b_y & b_z
\end{vmatrix}
\hat{i}
-
\begin{vmatrix}
a_x & a_z \\
b_x & b_z
\end{vmatrix}
\hat{j}
+
\begin{vmatrix}
a_x & a_y \\
b_x & b_y
\end{vmatrix}
\hat{k}
$$