AOS5 Topic 4: Cross Product

Vector Product

Geometric Definition of the Vector Product

Definition of the vector product: The vector product of $ \mathbf{a} $ and $ \mathbf{b} $ is denoted by $ \mathbf{a} \times \mathbf{b} $.

The magnitude of $ \mathbf{a} \times \mathbf{b} $ is equal to $ |\mathbf{a}| |\mathbf{b}| \sin \theta $, where $ \theta $ is the angle between $ \mathbf{a} $ and $ \mathbf{b} $.

The direction of $ \mathbf{a} \times \mathbf{b} $ is perpendicular to the plane containing $ \mathbf{a} $ and $ \mathbf{b} $, in the sense of the right-hand rule.

Note: The vector product is often called the cross product.

Watch the video below to understand the notes about the cross product more deeply.

Created with GeoGebra®, by Dave Nero, Link

The Magnitude of $ \mathbf{a} \times \mathbf{b} $

By definition, we have $ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \theta $ where $ \theta $ is the angle between $ \mathbf{a} $ and $ \mathbf{b} $.
The direction of $ \mathbf{a} \times \mathbf{b} $ using the right-hand rule:
Point your index finger along the vector $ \mathbf{a} $.
Point your middle finger along the vector $ \mathbf{b} $.
Keep your thumb at right angles to both $ \mathbf{a} $ and $ \mathbf{b} $. The direction of your thumb gives the direction of the vector $ \mathbf{a} \times \mathbf{b} $.

Calculating Cross Product

Created with GeoGebra®, by Lenore Horner, Link

The direction of $ \mathbf{a} \times \mathbf{b} $ using the right-hand rule

To find the direction of the vector $ \mathbf{a} \times \mathbf{b} $ using your right hand:

Point your index finger along the vector $ \mathbf{a} $.
Point your middle finger along the vector $ \mathbf{b} $.
Keep your thumb at right angles to both $ \mathbf{a} $ and $ \mathbf{b} $, as in the picture. The direction of your thumb gives the direction of the vector $ \mathbf{a} \times \mathbf{b} $.

The following two diagrams show $ \mathbf{a} \times \mathbf{b} $ and $ \mathbf{b} \times \mathbf{a} $.

The vector $ \mathbf{b} \times \mathbf{a} $ has the same magnitude as $ \mathbf{a} \times \mathbf{b} $, but the opposite direction. We can see that:

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

Thus, the vector product is not commutative.

Note: The vector product is also not associative. In general, we have $ (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} \neq \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) $.

Vector Product of Parallel Vectors

If $ \mathbf{a} $ and $ \mathbf{b} $ are parallel vectors, then $ \mathbf{a} \times \mathbf{b} = 0 $, since $ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin 0^\circ = 0 $.

Conversely, if $ \mathbf{a} $ and $ \mathbf{b} $ are non-zero vectors such that $ \mathbf{a} \times \mathbf{b} = 0 $, then $ \mathbf{a} $ and $ \mathbf{b} $ are parallel.

Vector Product of Perpendicular Vectors

If $ \mathbf{a} $ and $ \mathbf{b} $ are perpendicular vectors, then

\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin 90^\circ = |\mathbf{a}| |\mathbf{b}| \]

The three vectors $ \mathbf{a} $, $ \mathbf{b} $, and $ \mathbf{a} \times \mathbf{b} $ form a right-handed system of mutually perpendicular vectors,as shown in the diagram.

Vector Product in Component Form

Using the previous observations about the vector product of parallel and perpendicular vectors:

\[ \mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = 0 \]

\[ \mathbf{i} \times \mathbf{j} = \mathbf{k}, \quad \mathbf{j} \times \mathbf{k} = \mathbf{i}, \quad \mathbf{k} \times \mathbf{i} = \mathbf{j} \]

\[ \mathbf{j} \times \mathbf{i} = -\mathbf{k}, \quad \mathbf{k} \times \mathbf{j} = -\mathbf{i}, \quad \mathbf{i} \times \mathbf{k} = -\mathbf{j} \]

The diagram on the right may help you to follow the pattern among these vector products.

The vector product distributes over addition. That is:

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

These facts can be used to establish the following result.

Vector Product in Component Form

If $ \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} $ and $ \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} $, then

\[ \mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2) \mathbf{i} - (a_1 b_3 - a_3 b_1) \mathbf{j} + (a_1 b_2 - a_2 b_1) \mathbf{k} \]

Example 1

Find the vector product of $ \mathbf{a} = 3i + 3j + 8k $ and $ \mathbf{b} = i - 3j + 2k $,

and hence find a unit vector that is perpendicular to both $ \mathbf{a} $ and $ \mathbf{b} $.

The vector product can be evaluated as follows:

$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 3 & 8 \\ 1 & -3 & 2 \end{vmatrix} $
$= \mathbf{i} \begin{vmatrix} 3 & 8 \\ -3 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 8 \\ 1 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 3 \\ 1 & -3 \end{vmatrix} $
$= \mathbf{i} \left(3 \times 2 - 8 \times (-3)\right) - \mathbf{j} \left(3 \times 2 - 8 \times 1\right) + \mathbf{k} \left(3 \times (-3) - 3 \times 1\right) $
$= 30\mathbf{i} + 2\mathbf{j} - 12\mathbf{k} $

The magnitude of $ \mathbf{a} \times \mathbf{b} $ is $ \sqrt{30^2 + 2^2 + (-12)^2} = 2\sqrt{262} $.

Hence a unit vector perpendicular to both $ \mathbf{a} $ and $ \mathbf{b} $ is $ \frac{1}{{2\sqrt{262}}} (30i + 2j - 12k) $.

Example 2

Given that $ \mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{a} $, with $ \mathbf{a} \neq \mathbf{0} $.

Proof:

By assumption, we have

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

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

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

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

Since $ \mathbf{a} \neq \mathbf{0} $, it follows that either $ \mathbf{b} + \mathbf{c} = \mathbf{0} $ or the vectors $ \mathbf{a} $ and $ \mathbf{b} + \mathbf{c} $ are parallel.

Hence we must have $ \mathbf{b} = -\mathbf{c} $ or $ \mathbf{a} = k(\mathbf{b} + \mathbf{c}) $ for some $ k \in \mathbb{R} $.

Example 3

Find the angle between two vectors $ \mathbf{a} $ and $ \mathbf{b} $, where $ \mathbf{a} = \langle -4, 3, 0 \rangle $ and $ \mathbf{b} = \langle 2, 0, 0 \rangle $

Solution:

We know that the formula to find the angle between two vectors is:

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

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

Now, we have to find the cross product of two vectors $ \mathbf{a} $ and $ \mathbf{b} $:

\[ \mathbf{a} \times \mathbf{b} = \mathbf{i} \times (0) - \mathbf{j} \times (0) + \mathbf{k} \times (-6) = -6\mathbf{k} \]

While finding the angle between two vectors, substitute the magnitude of the vector value, thus:

\[ |\mathbf{a} \times \mathbf{b}| = 6 \]

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

\[ \theta = \sin^{-1}\left( \frac{6}{5 \cdot 2} \right) \]

\[ \theta = \sin^{-1}\left( \frac{3}{5} \right) = 36.87^\circ \]

Hence, the angle between two vectors, $ \mathbf{a} $ and $ \mathbf{b} $ ($ \theta $), is $ 36.87^\circ $.

Example 4

Simplify

$ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} + \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) $

To simplify $ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} + \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) $, we'll use the property that $ \mathbf{a} \times \mathbf{b} $ is perpendicular to $ \mathbf{a} $.

Given that $ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0 $ because $ \mathbf{a} \times \mathbf{b} $ is perpendicular to $ \mathbf{a} $, we have:

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

Since the dot product is commutative, $ \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} $, the expression simplifies to:

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

Example 5

Find a vector of magnitude 5 that is perpendicular to $ \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} $ and $ \mathbf{b} = \mathbf{i} - 2\mathbf{j} $.

Solution:

To find a vector of magnitude 5 that is perpendicular to $ \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} $ and $ \mathbf{b} = \mathbf{i} - 2\mathbf{j} $, we'll follow these steps:

Calculate the vector product $ \mathbf{c} = \mathbf{a} \times \mathbf{b} $.
Normalize $ \mathbf{c} $ by dividing it by its magnitude.
Scale $ \mathbf{c} $ to have a magnitude of 5.

1. Calculate the Vector Product $ \mathbf{c} = \mathbf{a} \times \mathbf{b} $

Given $ \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} $ and $ \mathbf{b} = \mathbf{i} - 2\mathbf{j} $, we compute:

$ \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -1 \\ 1 & -2 & 0 \end{vmatrix} $

$ = (\mathbf{j} \cdot 0 - \mathbf{k} \cdot (-2), -(\mathbf{i} \cdot 0 - \mathbf{k} \cdot 1), \mathbf{i} \cdot (-2) - \mathbf{j} \cdot 1) $

$ = (2, -1, -2) $

2. Normalize $ \mathbf{c} $

The magnitude of $ \mathbf{c} $ is $ |\mathbf{c}| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{9} = 3 $.

So, $ \mathbf{c} $ normalized is $ \frac{1}{3} (2\mathbf{i} - \mathbf{j} - 2\mathbf{k}) $.

3. Scale $ \mathbf{c} $ to Have a Magnitude of 5

To make the magnitude 5, we scale $ \mathbf{c} $ by $ \frac{5}{3} $.

Therefore, the vector of magnitude 5 that is perpendicular to $ \mathbf{a} $ and $ \mathbf{b} $ is $ \frac{5}{3}(2\mathbf{i} - \mathbf{j} - 2\mathbf{k}) = \frac{10}{3}\mathbf{i} - \frac{5}{3}\mathbf{j} - \frac{10}{3}\mathbf{k} $.

Example 6

Show that the area of a triangle formed by the position vectors $ \mathbf{a} $, $ \mathbf{b} $, and $ \mathbf{c} $ is given by $ \frac{1}{2} | \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} | $.

To show that the area of the triangle formed by the position vectors $ \mathbf{a} $, $ \mathbf{b} $, and $ \mathbf{c} $ is given by $ \frac{1}{2} | \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} | $, we can use the properties of the cross product and the formula for the area of a triangle.

Let's denote the area of the triangle formed by $ \mathbf{a} $, $ \mathbf{b} $, and $ \mathbf{c} $ as $ A $.

The area of a triangle formed by vectors $ \mathbf{u} $ and $ \mathbf{v} $ is half the magnitude of their cross product, i.e., $ A = \frac{1}{2} | \mathbf{u} \times \mathbf{v} | $.

Therefore, the area of the triangle formed by $ \mathbf{a} $, $ \mathbf{b} $, and $ \mathbf{c} $ can be written as:

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

Now, using the properties of the cross product:

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

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

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

We can rewrite the area $ A $ as:

$$ A = \frac{1}{2} (| \mathbf{a} \times \mathbf{b} | + | \mathbf{b} \times \mathbf{c} | + | \mathbf{c} \times \mathbf{a} |) + \frac{1}{2} (| -\mathbf{b} \times \mathbf{a} | + | -\mathbf{c} \times \mathbf{b} | + | -\mathbf{a} \times \mathbf{c} |) $$

$$ = \frac{1}{2} (| \mathbf{a} \times \mathbf{b} | + | \mathbf{b} \times \mathbf{c} | + | \mathbf{c} \times \mathbf{a} | + | \mathbf{b} \times \mathbf{a} | + | \mathbf{c} \times \mathbf{b} | + | \mathbf{a} \times \mathbf{c} |) $$

$$ = \frac{1}{2} (| \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} | + | \mathbf{b} \times \mathbf{a} + \mathbf{c} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} |) $$

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

Since $ | \mathbf{b} \times \mathbf{a} | = | \mathbf{c} \times \mathbf{b} | = | \mathbf{a} \times \mathbf{c} | $, we can rewrite the equation as:

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

Therefore, the area of the triangle formed by the position vectors $ \mathbf{a} $, $ \mathbf{b} $, and $ \mathbf{c} $ is given by $ \frac{1}{2} | \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} | $.

Example 7

A parallelogram $ OABC $ has one vertex at the origin $ O $ and two other vertices at the points $ A(0, 1, 3) $ and $ B(0, 2, 5) $. Find the area of $ OABC $.

Solution:

To find the area of the parallelogram $OABC$, we first need to find the area of the triangle formed by the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$. Let's calculate that first.

Given the vertices $O(0,0,0)$, $A(0,1,3)$, and $B(0,2,5)$, we can calculate the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ as follows:

\[
\overrightarrow{OA} = \langle 0 - 0, 1 - 0, 3 - 0 \rangle = \langle 0, 1, 3 \rangle
\]
\[
\overrightarrow{OB} = \langle 0 - 0, 2 - 0, 5 - 0 \rangle = \langle 0, 2, 5 \rangle
\]

Now, let's find the cross product of these vectors to get the area of the triangle $OAB$:

\[
\overrightarrow{OA} \times \overrightarrow{OB} = -\mathbf{i}
\]

The magnitude of $ \overrightarrow{OA} \times \overrightarrow{OB} $ is $ | \overrightarrow{OA} \times \overrightarrow{OB} | = 1 $. Therefore, the area of the triangle $ OAB $ is $ \frac{1}{2} \times 1 = \frac{1}{2} $.

Now, since the area of a parallelogram is twice the area of a triangle formed by its adjacent sides, the area of the parallelogram $ OABC $ is $ 2 \times \frac{1}{2} = 1 $.

So, the area of the parallelogram $ OABC $ is $ 1 $.

Exercise 1

Practice Makes Perfect

Exercise 2

Practice Makes Perfect

Exercise 3

Practice Makes Perfect

Exercise 4

Practice Makes Perfect

Exercise 5

Practice Makes Perfect

AOS5 Topic 4: Cross Product

Vector Product

Geometric Definition of the Vector Product

The Magnitude of \( \mathbf{a} \times \mathbf{b} \)

Calculating Cross Product

The direction of \( \mathbf{a} \times \mathbf{b} \) using the right-hand rule

Vector Product of Parallel Vectors

Vector Product of Perpendicular Vectors

Vector Product in Component Form

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5