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Cross product of $$$\left\langle -2, 3, 1\right\rangle$$$ and $$$\left\langle 7, -4, 0\right\rangle$$$
Your Input
Calculate $$$\left\langle -2, 3, 1\right\rangle\times \left\langle 7, -4, 0\right\rangle$$$.
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: $$$\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\-2 & 3 & 1\\7 & -4 & 0\end{array}\right|$$$.
Now, just expand along the first row (for steps in finding a determinant, see determinant calculator):
$$$\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\-2 & 3 & 1\\7 & -4 & 0\end{array}\right| = \left|\begin{array}{cc}3 & 1\\-4 & 0\end{array}\right| \mathbf{\vec{i}} - \left|\begin{array}{cc}-2 & 1\\7 & 0\end{array}\right| \mathbf{\vec{j}} + \left|\begin{array}{cc}-2 & 3\\7 & -4\end{array}\right| \mathbf{\vec{k}} = \left(\left(3\right)\cdot \left(0\right) - \left(1\right)\cdot \left(-4\right)\right) \mathbf{\vec{i}} - \left(\left(-2\right)\cdot \left(0\right) - \left(1\right)\cdot \left(7\right)\right) \mathbf{\vec{j}} + \left(\left(-2\right)\cdot \left(-4\right) - \left(3\right)\cdot \left(7\right)\right) \mathbf{\vec{k}} = 4 \mathbf{\vec{i}} + 7 \mathbf{\vec{j}} - 13 \mathbf{\vec{k}}$$$
Thus, $$$\left\langle -2, 3, 1\right\rangle\times \left\langle 7, -4, 0\right\rangle = \left\langle 4, 7, -13\right\rangle.$$$
Answer
$$$\left\langle -2, 3, 1\right\rangle\times \left\langle 7, -4, 0\right\rangle = \left\langle 4, 7, -13\right\rangle$$$A