Cross product of $$$\left\langle 7, -4, 0\right\rangle$$$ and $$$\left\langle -3, 2, 1\right\rangle$$$

The calculator will find the cross product of two vectors $$$\left\langle 7, -4, 0\right\rangle$$$ and $$$\left\langle -3, 2, 1\right\rangle$$$, with steps shown.
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Your Input

Calculate $$$\left\langle 7, -4, 0\right\rangle\times \left\langle -3, 2, 1\right\rangle$$$.

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: $$$\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\7 & -4 & 0\\-3 & 2 & 1\end{array}\right|$$$.

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

$$$\left|\begin{array}{ccc}\mathbf{\vec{i}} & \mathbf{\vec{j}} & \mathbf{\vec{k}}\\7 & -4 & 0\\-3 & 2 & 1\end{array}\right| = \left|\begin{array}{cc}-4 & 0\\2 & 1\end{array}\right| \mathbf{\vec{i}} - \left|\begin{array}{cc}7 & 0\\-3 & 1\end{array}\right| \mathbf{\vec{j}} + \left|\begin{array}{cc}7 & -4\\-3 & 2\end{array}\right| \mathbf{\vec{k}} = \left(\left(-4\right)\cdot \left(1\right) - \left(0\right)\cdot \left(2\right)\right) \mathbf{\vec{i}} - \left(\left(7\right)\cdot \left(1\right) - \left(0\right)\cdot \left(-3\right)\right) \mathbf{\vec{j}} + \left(\left(7\right)\cdot \left(2\right) - \left(-4\right)\cdot \left(-3\right)\right) \mathbf{\vec{k}} = - 4 \mathbf{\vec{i}} - 7 \mathbf{\vec{j}} + 2 \mathbf{\vec{k}}$$$

Thus, $$$\left\langle 7, -4, 0\right\rangle\times \left\langle -3, 2, 1\right\rangle = \left\langle -4, -7, 2\right\rangle.$$$

Answer

$$$\left\langle 7, -4, 0\right\rangle\times \left\langle -3, 2, 1\right\rangle = \left\langle -4, -7, 2\right\rangle$$$A