Cross Product (Geometric)
Vector multiplication can be completed using the dot product or cross product. The result of the cross product is a vector. We can calculate the cross product using geometric or algebraic vectors.
The geometric cross product of two vectors \( \vec{a}\) and \(\vec{b}\) is defined by:
\(\vec{a}\times\vec{b} = \vert\vec{a}\vert\vert\vec{b}\vert\sin{\theta}\)
Where \(\theta\) represents the angle between the tails.
The direction of the cross product vector, \( \hat{n} \), is the normal vector perpendicular to both \( \vec{a}\) and \(\vec{b}\). You can determine the direction of the normal using the right hand rule: Point your right index finger in the direction of the first vector, point your right middle finger in the direction of the second vector. The direction of your thumb is the direction of the normal vector.
Cross Product Properties
| Name | Property |
|---|---|
| Anti-Commutative | \(\vec{a}\times\vec{b} = -\vec{b}\times\vec{a} \) |
| Distributive | \( \vec{a}\times(\vec{b}+c)=\vec{a}\times\vec{b}+\vec{a}\times\vec{c} \) |
| Multiplication by a Constant | \( k(\vec{a}\times\vec{b}) = \vec{a}\times k\vec{b} = k\vec{a}\times\vec{b} \) |
| Non-Associative | \( (\vec{a}\times\vec{b})\times\vec{c} \ne \vec{a}\times(\vec{b}\times\vec{c}) \) |
| Triple Scalar Product | \( \vec{u}\cdot(v\times\vec{w}) = (\vec{u}\times\vec{v})\cdot\vec{w} \) |
| Zero Cross Product | \(\vec{a}\times\vec{0} = 0\) |
| Collinear Vectors | \(\vec{a}\times\vec{b} = 0\) |
| Cross Product of a Vector with Itself | \(\vec{a}\times\vec{a} = 0\) |
Find \(\vert\vec{u}\times\vec{v}\vert \) for each of the following pairs of vectors. State whether \(\vec{u}\times\vec{v} \) is directed into or out of the page.
\(\vert\vec{u}\vert = 12, \; \vert\vec{v}\vert = 5, \; \theta=68^{\circ}\)
First, we can sketch a diagram of the vectors with the angle between them:
Next, we can use cross product to determine the magnitude and direction:
\(\vert\vec{u}\times\vec{v}\vert = \vert\vec{u}\vert\vert\vec{v}\vert\sin{\theta}\)
\(= (12)(5)\sin{68^{\circ}}\)
\(= 55.6\)
Therefore, we can determine that the cross product is \(\boldsymbol{55.6 \; [\textbf{units}]}\) directed into the page.
\(\vert\vec{u}\vert = 18, \; \vert\vec{v}\vert = 25, \; \theta=120^{\circ}\)
First, we can sketch a diagram of the vectors with the angle between them:
Next, we can use cross product to determine the magnitude and direction:
\(\vert\vec{u}\times\vec{v}\vert = \vert\vec{u}\vert\vert\vec{v}\vert\sin{\theta}\)
\(= (18)(25)\sin{120^{\circ}}\)
\(= 450\left(\dfrac{\sqrt{3}}{2}\right)\)
\(= 225 \sqrt{3}\)
Therefore, we can determine that the cross product is \(\boldsymbol{225 \sqrt{3} \; [\textbf{units}]}\) directed out of the page.
Categorizing Vectors
Try and fit the following derivatives into their correct description.
\(\vec{a}\cdot(\vec{b}\times\vec{c})\)
\( \vec{a}\times(\vec{b}\cdot\vec{c})\)
\(\vec{a}\cdot(\vec{b}\cdot\vec{c})\)
\(\vec{a}\times(\vec{b}\times\vec{c})\)
\((\vec{a}\cdot\vec{b})\times(\vec{b}\cdot\vec{c})\)
\((\vec{a}\times\vec{b})\cdot(\vec{b}\times\vec{c})\)
\((\vec{a}\times\vec{b})+(\vec{b}\times\vec{c})\)
\((\vec{a}\cdot\vec{b})+(\vec{b}\cdot\vec{c})\)
\((\vec{a}+\vec{b})\cdot\vec{c}\)
\((\vec{a}+\vec{b})\times\vec{c}\)
\((\vec{a}\times\vec{b})-\vec{c}\)
\((\vec{a}\cdot\vec{b})-\vec{c}\)