Crossing Two Vectors

Solution 1: $\begin{aligned} \vec{u} \times \vec{v} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 9 & 0 \\ 13 & 18 & 0 \end{vmatrix} \\ &= \mathbf{i}(9\times 0 - 18\times 0) + \mathbf{j}(2\times 0 - 13\times 0) + \mathbf{k}(2\times 18 - 9\times 13) \\ &= -81\mathbf{k} \\ \end{aligned}$

Hence $|\vec{u} \times \vec{v}| = |-81| = 81$ .

Solution 2: From the cross product formula, we have $\vec{u} \times \vec{v} = |\vec{u}||\vec{v}|\sin \theta$ . We can easily calculate the magnitude of the vectors $|\vec{u}| = \sqrt{2 ^2 + 9 ^2} = \sqrt{85}$ and $|\vec{v}| = \sqrt{13 ^2 + 18 ^2} = \sqrt{493}$ .

We calculate $\sin \theta$ by using the dot product formula $\cos \theta = \frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$ and substituting $\cos \theta = \sqrt{1 - \sin^2 \theta}$ into the formula. So, we have $\begin{aligned} \sqrt{1-\sin^2\theta} &= \frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|} \\ &= \frac{2\cdot13 + 9\cdot 18}{\sqrt{2 ^2 + 9 ^2}\sqrt{13 ^2 + 18 ^2}} \\ \sin^2\theta &= 1 - \frac{(2\cdot 13 + 9\cdot 18) ^2}{(2 ^2 + 9 ^2)(13 ^2 + 18 ^2)} \\ &= \frac{ 2 ^2 \cdot 18 ^2 + 9 ^2 \cdot 13 ^2 - 2 \cdot 2 \cdot 13 \cdot 9 \cdot 18}{(2 ^2 + 9 ^2)(13 ^2 + 18 ^2)} \\ &= \frac{(2\times 18 - 9 \times 13)^2}{(2 ^2 + 9 ^2)(13 ^2 + 18 ^2)} \\ |\sin \theta| & = \frac{81}{\sqrt{85} \cdot \sqrt{493}} \\ \end{aligned}$

Hence $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}||\sin \theta| = \sqrt{85} \cdot \sqrt{493} \cdot \frac{81}{\sqrt{85} \cdot \sqrt{493}} = 81$ .