# Cross Product Calculator

The online calculator will find the cross product of two vectors, with steps shown.

$\langle$ $\rangle$
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$\langle$ $\rangle$
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Calculate $\left\langle 3, 1, 4\right\rangle\times \left\langle -2, 0, 5\right\rangle$.

## 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: $\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\3 & 1 & 4\\-2 & 0 & 5\end{array}\right|$.

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

$\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\3 & 1 & 4\\-2 & 0 & 5\end{array}\right| = \left|\begin{array}{cc}1 & 4\\0 & 5\end{array}\right| \mathbf{\vec{i}} - \left|\begin{array}{cc}3 & 4\\-2 & 5\end{array}\right| \mathbf{\vec{j}} + \left|\begin{array}{cc}3 & 1\\-2 & 0\end{array}\right| \mathbf{\vec{k}} = \left(\left(1\right)\cdot \left(5\right) - \left(4\right)\cdot \left(0\right)\right) \mathbf{\vec{i}} - \left(\left(3\right)\cdot \left(5\right) - \left(4\right)\cdot \left(-2\right)\right) \mathbf{\vec{j}} + \left(\left(3\right)\cdot \left(0\right) - \left(1\right)\cdot \left(-2\right)\right) \mathbf{\vec{k}} = 5 \mathbf{\vec{i}} - 23 \mathbf{\vec{j}} + 2 \mathbf{\vec{k}}$

Thus, $\left\langle 3, 1, 4\right\rangle\times \left\langle -2, 0, 5\right\rangle = \left\langle 5, -23, 2\right\rangle.$

$\left\langle 3, 1, 4\right\rangle\times \left\langle -2, 0, 5\right\rangle = \left\langle 5, -23, 2\right\rangle$A