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

The calculator will find the cross product of two vectors $\left\langle 1, 3, 2 t\right\rangle$ and $\left\langle 0, 0, 2\right\rangle$, with steps shown.
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Calculate $\left\langle 1, 3, 2 t\right\rangle\times \left\langle 0, 0, 2\right\rangle$.

### Solution

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