Cross product of $$$\left\langle 1, 3, 2 t\right\rangle$$$ and $$$\left\langle 0, 0, 2\right\rangle$$$

Calculate $$$\left\langle 1, 3, 2 t\right\rangle\times \left\langle 0, 0, 2\right\rangle$$$.

To find the cross product, we form a formal determinant the first row of which consists of unit vectors, the second row is our first vector, and the third row is our second vector: $$$\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\1 & 3 & 2 t\\0 & 0 & 2\end{array}\right|$$$.

Now, just expand along the first row (for steps in finding a determinant, see determinant calculator):

$$$\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\1 & 3 & 2 t\\0 & 0 & 2\end{array}\right| = \left|\begin{array}{cc}3 & 2 t\\0 & 2\end{array}\right| \mathbf{\vec{i}} - \left|\begin{array}{cc}1 & 2 t\\0 & 2\end{array}\right| \mathbf{\vec{j}} + \left|\begin{array}{cc}1 & 3\\0 & 0\end{array}\right| \mathbf{\vec{k}} = \left(\left(3\right)\cdot \left(2\right) - \left(2 t\right)\cdot \left(0\right)\right) \mathbf{\vec{i}} - \left(\left(1\right)\cdot \left(2\right) - \left(2 t\right)\cdot \left(0\right)\right) \mathbf{\vec{j}} + \left(\left(1\right)\cdot \left(0\right) - \left(3\right)\cdot \left(0\right)\right) \mathbf{\vec{k}} = 6 \mathbf{\vec{i}} - 2 \mathbf{\vec{j}} + 0$$$

Thus, $$$\left\langle 1, 3, 2 t\right\rangle\times \left\langle 0, 0, 2\right\rangle = \left\langle 6, -2, 0\right\rangle.$$$

$$$\left\langle 1, 3, 2 t\right\rangle\times \left\langle 0, 0, 2\right\rangle = \left\langle 6, -2, 0\right\rangle$$$A