Considering vectors in \(3\) dimensions allows for the possibility of a vector being mutually orthogonal to two other non-collinear vectors. This is not possible in \(2\) dimensions.

If either \(\vec{a}\) or \(\vec{b}\) is \(\vec{0}\) or if \(\vec{a}\) and \(\vec{b}\) are collinear, then their cross product produces \(\vec{0}\).

The cross product is sometimes referred to as the vector product.

Given any two vectors \(\vec{a}=(a_1,a_2,a_3)\) and \(\vec{b}=(b_1,b_2,b_3)\), find a vector that is orthogonal to both \(\vec{a}\) and \(\vec{b}\).

Let the required vector be \((x,y,z)\).

If \(\vec{a}\) is orthogonal to \((x,y,z)\), then

If \(\vec{b}\) is orthogonal to \((x,y,z)\), then

Eliminate \(z\) from the two equations:

Subtracting \((4)-(3)\) gives

If we repeat the same procedure, instead multiplying equation \((1)\) by \(b_1\) and equation \((2)\) by \(a_1\), we may eliminate \(x\) to get

Therefore, we have

for some \(c\in\mathbb{R}\).

Finally, we see that

Since any scalar multiple of \((x,y,z)\) will be orthogonal to both \(\vec{a}\) and \(\vec{b}\) (you can verify this by taking the dot product), we may simplify the vector above by letting \(c=1\).

In fact, by definition, the cross product of \(\vec{a}=(a_1,a_2,a_3)\) and \(\vec{b}=(b_1,b_2,b_3)\) is exactly the vector given above when \(c=1\).

It is worth noting that, by definition, the vector, \(\vec{b}\times \vec{a}\), is determined by letting \(c=-1\) so that

Determine a vector orthogonal to \(\vec{a}= (\underset{a_1}{1}, \underset{a_2}{2}, \underset{a_3}{-4})\) and \(\vec{b}=(\underset{b_1}{3}, \underset{b_2}{-1}, \underset{b_3}{5}) \).

Solution

Using the cross product formula,

To check our work, we take the dot product of our result with each of \(\vec{a}\) and \(\vec{b}\):