Area of a Parallelogram spanned by Vectors

The cross product of two vectors can be determined in any of the following two ways :

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

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

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

• \(|\vec{a} \times \vec{b}| = |\vec{b} \times \vec{a}| = |\vec{a}| \times |\vec{b}| \times \sin{\theta}\) ……where \(\theta\) is the angle between the vectors.

The first result gives the cross product in component form. The second result gives the magnitude of the cross product. 

An interesting application of cross product occurs in geometry. The magnitude of \(\vec{a} \times \vec{b}\) is actually the area of the parallelogram spanned by \(\vec{a}\) and \(\vec{b}\).

Let’s see this through an example.

A parallelogram is a four sided 2D shape with opposite sides parallel and equal. 

Question :-

\(\vec{a} = 1\hat{i} + 1\hat{j} + 4\hat{k}\)

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

Find the area of the parallelogram spanned by \(\vec{a}\) and \(\vec{b}\).

Since it is said that the given vectors are forming the parallelogram, it is indirectly implied that they are not parallel vectors, otherwise the parallelogram will not exist. Let \(\theta\) be the angle between the given vectors. 

Taking these two vectors as the non-parallel adjacent sides of the parallelogram, we can define the following two vectors to complete the parallelogram : 

To complete the parallelogram we will define a vector parallel(or anti-parallel) to \(\vec{a}\) and having the same magnitude(length) as \(\vec{a}\). And also a vector parallel(or anti-parallel) to \(\vec{b}\) and having the same length as \(\vec{b}\)(see the right image). 

The area of a parallelogram is the product of its base and its height. 

If we take \(\vec{a}\) as its base, the height(h) can simply be determined using the trigonometric sine function. 

From the figure, 

\(\sin{\theta} = \frac{h}{|\vec{b}|}\)

\(\implies h = |\vec{b}| \sin{\theta}\)

So the area of the parallelogram is :

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

But notice the term on the right is nothing but the magnitude of \(\vec{a} \times \vec{b}\).

So as you can see, the magnitude of the cross product of two vectors is the area of the parallelogram spanned by the vectors. 

Now to find this area, we can either find the magnitudes of \(\vec{a}\) and \(\vec{b}\) separately from their given components, then find the angle between the vectors through other means. And finally substitute all of that in the above relation. This may be a tedious process.

OR, we can simply get \(|\vec{a} \times \vec{b}|\) by first finding \(\vec{a} \times \vec{b}\) in component form, since the given vectors are in component form too. 

\(\vec{a} \times \vec{b} = \Biggl|\matrix{\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 4 \\ 2 & 3 & -1}\Biggl|\)

\( = (-1 - 12)\hat{i} - (-1 - 8)\hat{j} + (3 - 2)\hat{k}\)

\( = -13\hat{i} + 9\hat{j} + 1\hat{k}\)

\(\implies |\vec{a} \times \vec{b}| = \sqrt{(-13)^2 + 9^2 + 1^2}\)

            \( = \sqrt{251}\)

And so the area of the parallelogram spanned by \(\vec{a}\) and \(\vec{b}\) is \(\sqrt{251}\) sq.units.