# Math Insight

### Cross product examples

#### Example 1

Calculate the cross product between $\vc{a} = (3, -3, 1)$ and $\vc{b} = (4,9,2)$.

Solution: The cross product is \begin{align*} \vc{a} \times \vc{b} &= \left| \begin{array}{ccc} \vc{i} & \vc{j} & \vc{k}\\ 3 & -3 & 1\\ 4 & 9 & 2 \end{array} \right|\\ &= \vc{i} (-3\cdot 2 -1 \cdot 9) - \vc{j}(3\cdot 2- 1 \cdot 4) + \vc{k}(3 \cdot 9 + 3 \cdot 4)\\ &= -15 \vc{i} -2 \vc{j} + 39 \vc{k} \end{align*}

#### Example 2

Calculate the area of the parallelogram spanned by the vectors $\vc{a} = (3, -3, 1)$ and $\vc{b} = (4,9,2)$.

Solution: The area is $\| \vc{a} \times \vc{b}\|$. Using the above expression for the cross product, we find that the area is $\sqrt{15^2+2^2+39^2} = 5 \sqrt{70}$.

#### Example 3

Calculate the area of the parallelogram spanned by the vectors $\vc{a} = (3,-3,1)$ and $\vc{c} = (-12, 12, -4)$.

Solution: \begin{align*} \vc{a} \times \vc{c} &= \left| \begin{array}{ccc} \vc{i} & \vc{j} & \vc{k}\\ 3 & -3 & 1\\ -12 & 12 & -4 \end{array} \right|\\ &= (0,0,0) \end{align*}

The magnitude of the zero vector is zero, so the area of the parallelogram is zero. What happened?