Cross product of $$$\left\langle 2 t, 3 t^{2}, 1\right\rangle$$$ and $$$\left\langle 2, 6 t, 0\right\rangle$$$

Calculate $$$\left\langle 2 t, 3 t^{2}, 1\right\rangle\times \left\langle 2, 6 t, 0\right\rangle$$$.

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: $$$\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\2 t & 3 t^{2} & 1\\2 & 6 t & 0\end{array}\right|$$$.

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

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

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

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