The formula derived earlier, while useful, can be tedious to memorize. There is an alternative method for computing the cross product that involves writing the coordinates of \(\vec{a}=(a_1,a_2,a_3)\) and \(\vec{b}=(b_1,b_2,b_3)\) in an array as shown.

Cross out the first and last columns.

Proceed using the “down minus up” method. First draw a “down” arrow from \(a_2\) to \(b_3\) and determine the “down” product \(a_2b_3\).

Then draw an “up” arrow from \(b_2\) to \(a_3\) and compute the “up” product \(a_3b_2\). The \(x\) component of the cross product is the value of the down product minus the up product, \(a_2b_3-a_3b_2\).

To determine the \(y\) component of the cross product, repeat with \(a_3b_1\) and \(a_1b_3\).

Finally, to determine the \(z\) component of the cross product repeat using \(a_1b_2\) and \(a_2b_1\).

This is the original cross product formula!

Applying this method to example 1, with \((a_1,a_2,a_3)=(1,2,-4)\) and \((b_1,b_2,b_3)=(3,-1,5)\), we get the following.

and so

as expected.

Find two vectors orthogonal to both \(\vec{a}=(1,-3,1)\)and \(\vec{b}=(0,4,-1)\).

Solution

Setting up the array to compute the cross product

gives

Check:

Therefore, one vector orthogonal to both \(\vec{a}\) and \(\vec{b}\) is \((-1,1,4)\).

Taking a scalar multiple of this vector, (say, its opposite vector) \((1,-1,-4)\) is sufficient for finding a second.