Cross products of two vectors

Vector product of two non-parallel vectors is a vector perpendicular to both the vectors. If \(\vec{a}\) and \(\vec{b}\) are non-parallel vectors then \(\vec{a} \times \vec{b}\) is their cross product - a vector making 90 degree angle with both \(\vec{a}\) and \(\vec{b}\). It is also known as cross product.

Another way to think of it is to imagine a plane “containing” the two non-parallel vectors. The vector product makes a 90 degree angle with the plane containing them, i.e. it is a vector normal to the plane. 

Let \(\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}\) and,

      \(\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}\)  

Then, \(\vec{a} \times \vec{b}\) is equal to the following determinant :

\(\vec{a} \times \vec{b} = \Biggl|\matrix{\hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}}\Biggl|\)

The first row are the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) along the three axes. The second row are the three components of \(\vec{a}\)(the first vector from the left in the cross product). And the third row are the three components of \(\vec{b}\).

\(\vec{a} \times \vec{b} = (a_{2}b_{3} - b_{2}a_{3}) \hat{i} - (a_{1}b_{3} - b_{1}a_{3}) \hat{j} + (a_{1}b_{2} - b_{1}a_{2}) \hat{k}\)

This is a vector product in component form i.e., in terms of \(\hat{i}, \hat{j}, \hat{k}\). 

The magnitude of \(\vec{a} \times \vec{b}\) is just like how we find the magnitude of any vector given in component form (By adding the squares of the components and taking the square root of the entire term).

\(|\vec{a} \times \vec{b}| = \sqrt{(a_{2}b_{3} - b_{2}a_{3})^{2} + (a_{1}b_{3} - b_{1}a_{3})^{2} + (a_{1}b_{2} - b_{1}a_{2})^{2}}\)

Like \(\vec{a} \times \vec{b}\), \(\vec{b} \times \vec{a}\) is also a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\). It has the same magnitude as \(\vec{a} \times \vec{b}\) but points in the direction opposite to it. In other words, \(\vec{b} \times \vec{a}\) is anti-parallel to \(\vec{a} \times \vec{b}\). 

\(\vec{a} \times \vec{b} = - (\vec{b} \times \vec{a})\)

For \(\vec{b} \times \vec{a}\), simply switch the places of the components of \(\vec{a}\) and \(\vec{b}\) in the second and third rows in the determinant. 

\(\vec{b} \times \vec{a} = \Biggl|\matrix{\hat{i} & \hat{j} & \hat{k} \\ b_{1} & b_{2} & b_{3} \\ a_{1} & a_{2} & a_{3}}\Biggl|\)

\(\vec{b} \times \vec{a} = (b_{2}a_{3} - a_{2}b_{3}) \hat{i} - (b_{1}a_{3} - a_{1}b_{3}) \hat{j} + (b_{1}a_{2} - a_{1}b_{2}) \hat{k}\)

Whichever vector is on the left will have its components in the second row. In this case, \(\vec{b}\) is on the left, so we have \(b_{1}; b_{2}; b_{3}\) in the second row. 

There is also another way to determine the magnitude of a vector product. The magnitude of \(\vec{a} \times \vec{b}\) is the magnitude of \(\vec{a}\)…times the magnitude of \(\vec{b}\)…times \(\sin{\theta}\), where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\) as shown in the figure above. 

\(|\vec{a} \times \vec{b}| = |\vec{b} \times \vec{a}| = |\vec{a}| \times |\vec{b}| \times \sin{\theta}\)

So far we have looked at two ways of determining the magnitude of a vector product. But about its direction, I have only told you that it is at right angles to both the vectors and the plane containing them. For any vector normal to a plane there are only two possible directions. \(\vec{a} \times \vec{b}\) will point in one of them and \(\vec{b} \times \vec{a}\) in the other. There is a right-hand rule which tells us which one will point in which direction. You can find out more about this rule here. 

What if \(\vec{a}\) and \(\vec{b}\) are parallel?

In that case, the vector product equals zero vector. We can quickly check this out. If \(\vec{a}\) and \(\vec{b}\) are parallel then the angle between them is 0 or 180 degrees. 

\(\sin{0^\circ} = \sin{180^\circ} = 0\)

\(\implies \vec{a} \times \vec{b} = \vec{b} \times \vec{a} = \vec{0}\)

The converse is also true. That is, if the cross product of two vectors is \(\vec{0}\) then the vectors are parallel. Infact, the converse is used quite extensively in problems to help determine if the vectors are parallel from their components. If the determinant yields \(\vec{0}\) then they are parallel vectors. If the determinant is a non-zero vector then the vectors are not parallel.