Areas from the cross product

What you will be able to do

• find the area of a parallelogram and of a triangle from vectors or from vertices;
• test whether three points lie on one line, with the cross product;
• use the identity $\lvert\vec a\times\vec b\rvert^2 = \lvert\vec a\rvert^2\lvert\vec b\rvert^2 - (\vec a\cdot\vec b)^2$.

A plot of land

A surveyor knows the position of three corner posts, with heights, and wants the area of the triangle of land. Measuring the sides and using a formula takes time. With vectors the area comes out of one cross product.

Area of a parallelogram and a triangle

Draw the parallelogram with the sides $\vec a$ and $\vec b$. Take $\vec a$ as the base. The height is $h = \lvert\vec b\rvert\sin\theta$. So the area is base times height, which is $\lvert\vec a\rvert\lvert\vec b\rvert\sin\theta$:

A diagonal cuts the parallelogram into two equal triangles, so

For a triangle with vertices $A$, $B$, $C$, use the two sides from one vertex:

Other forms follow. A parallelogram with the diagonals $\vec d_1$ and $\vec d_2$ has the area $\dfrac{1}{2}\lvert\vec d_1\times\vec d_2\rvert$. The same holds for any quadrilateral in a plane. The height of a triangle on the base $AB$ is $\dfrac{2\times\text{area}}{\lvert\overrightarrow{AB}\rvert}$.

Three points on a line

When $A$, $B$, $C$ lie on one line, the triangle has no area, so $\overrightarrow{AB}\times\overrightarrow{AC} = \vec 0$. This gives a test for collinear points that works in space and has no division. The vectors are parallel exactly when the cross product is zero.

The identity of Lagrange

Since $\sin^2\theta = 1 - \cos^2\theta$,

It gives the cross product's size from the dot product, and the dot product's size from the cross product. It also shows that $\lvert\vec a\times\vec b\rvert^2 + (\vec a\cdot\vec b)^2 = \lvert\vec a\rvert^2\lvert\vec b\rvert^2$, which is $\sin^2 + \cos^2 = 1$ in disguise.

A worked example

1. A parallelogram. The sides are $\vec a = 3\hat i + 4\hat k$ and $\vec b = 5\hat j$. The cross product is $(0 - 20)\hat i - (0 - 0)\hat j + (15 - 0)\hat k = -20\hat i + 15\hat k$, with size $\sqrt{400 + 225} = 25$. The triangle on $\vec a$ and $\vec b$ has area $12.5$.
2. A triangle from vertices. $A(1, 2, 3)$, $B(2, 4, 5)$ and $C(4, 3, 3)$ give $\overrightarrow{AB} = (1, 2, 2)$ and $\overrightarrow{AC} = (3, 1, 0)$. Then $\overrightarrow{AB}\times\overrightarrow{AC} = (-2, 6, -5)$ with size $\sqrt{65}$. The area is $\dfrac{\sqrt{65}}{2} \approx 4.03$.
3. Lagrange. Let $\lvert\vec a\rvert = 4$, $\lvert\vec b\rvert = 5$ and $\vec a\cdot\vec b = 12$. Then $\lvert\vec a\times\vec b\rvert^2 = 400 - 144 = 256$, so $\lvert\vec a\times\vec b\rvert = 16$ and $\sin\theta = \dfrac{16}{20} = 0.8$.
4. Collinear points. $P(0, 1, -1)$, $Q(2, 5, 3)$ and $R(5, 11, 9)$ give $\overrightarrow{PQ} = (2, 4, 4)$ and $\overrightarrow{PR} = (5, 10, 10)$. These are parallel, so the cross product is zero and the points are collinear.

Practise with new numbers

Try it in Python

The program finds the area of the triangle in the worked example and tests three collinear points.

Common mistakes

• Forgetting the half for a triangle.
• Using the dot product for area. The area needs the sine, so it is the cross product.
• Taking the sides from different vertices. Use $\overrightarrow{AB}$ and $\overrightarrow{AC}$, both from $A$.
• Reading a zero cross product as a zero vector. It means the vectors are parallel, or one is zero.
• A negative area. Take the size of the cross product, which is never negative.

Remember these

• Parallelogram: $\lvert\vec a\times\vec b\rvert$. Triangle: $\dfrac{1}{2}\lvert\overrightarrow{AB}\times\overrightarrow{AC}\rvert$. Diagonals: $\dfrac{1}{2}\lvert\vec d_1\times\vec d_2\rvert$.
• Collinear points give $\overrightarrow{AB}\times\overrightarrow{AC} = \vec 0$.
• $\lvert\vec a\times\vec b\rvert^2 = \lvert\vec a\rvert^2\lvert\vec b\rvert^2 - (\vec a\cdot\vec b)^2$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the area of the parallelogram with the sides $\hat i + \hat j$ and $\hat j + \hat k$.

2. Find the area of the triangle with the vertices $(0, 0, 0)$, $(2, 0, 0)$ and $(0, 3, 4)$.

3. Two vectors have size $10$ each, and $\vec a\cdot\vec b = 60$. Find $\lvert\vec a\times\vec b\rvert$.

4. Are $P(1, 1, 1)$, $Q(2, 3, 4)$ and $R(4, 7, 10)$ collinear?

5. Can the size of $\vec a\times\vec b$ be larger than $\lvert\vec a\rvert\lvert\vec b\rvert$?

References

• NCERT, Mathematics Part II for Class XII, Chapter 10, section on the vector product and its applications to area.
• OpenStax, Calculus Volume 3, Chapter 2, section on the cross product.