Parallelogram Area from the Vector Product

The parallelogram two vectors span

Draw two vectors a and b from the same corner. Draw a again from the head of b, and b again from the head of a: the two meet at the fourth corner, the head of a + b. The four sides make a parallelogram, the one a and b span.

The area of a parallelogram is its base times its perpendicular height. Take a as the base, so the base is |a|. The height is the distance from the head of b straight down to the line of a. In the right triangle with hypotenuse |b| and the angle $\theta$ between a and b, that height is the side opposite $\theta$, $|b| \sin \theta$.

So the area of the parallelogram is $|a| |b| \sin \theta$.

The base is |a|, and the perpendicular height, measured outside the slanted side b, is $|b| \sin \theta$.

The length of the vector product

The length of a × b is exactly that area: $|a \times b| = |a| |b| \sin \theta$. So the vector product carries the parallelogram in its length, as well as a direction at right angles to it.

Here is why, for two flat vectors $a = \left(a_{1}, a_{2}, 0\right)$ and $b = \left(b_{1}, b_{2}, 0\right)$. Their vector product is $\left(0, 0, a_{1}b_{2} - a_{2}b_{1}\right)$, so $|a \times b|^{2} = \left(a_{1}b_{2} - a_{2}b_{1}\right)^{2}$. Now work out $|a|^{2} |b|^{2} - \left(a \cdot b\right)^{2} = \left(a_{1}^{2} + a_{2}^{2}\right)\left(b_{1}^{2} + b_{2}^{2}\right) - \left(a_{1}b_{1} + a_{2}b_{2}\right)^{2}$. The terms $a_{1}^{2}b_{1}^{2}$ and $a_{2}^{2}b_{2}^{2}$ cancel, leaving $a_{1}^{2}b_{2}^{2} + a_{2}^{2}b_{1}^{2} - 2a_{1}a_{2}b_{1}b_{2}$, which is $\left(a_{1}b_{2} - a_{2}b_{1}\right)^{2}$. So $|a \times b|^{2} = |a|^{2} |b|^{2} - \left(a \cdot b\right)^{2}$. With three components there are more terms, and they cancel in the same way.

Put in $a \cdot b = |a| |b| \cos \theta$: $|a \times b|^{2} = |a|^{2} |b|^{2} - |a|^{2} |b|^{2} \cos^{2}\theta = |a|^{2} |b|^{2} \left(1 - \cos^{2}\theta\right)$. Since $\sin^{2}\theta + \cos^{2}\theta = 1$, that is $|a|^{2} |b|^{2} \sin^{2}\theta$. For $\theta$ from 0° to 180°, $\sin \theta$ is not negative, so $|a \times b| = |a| |b| \sin \theta$.

Checked on numbers

Take a = (2, 1, 0) and b = (1, 3, 0). Then a × b = (0, 0, 2 × 3 − 1 × 1) = (0, 0, 5), whose length is 5.

The other way: $|a| = \sqrt{5}$, $|b| = \sqrt{10}$ and a · b = 2 + 3 = 5, so $\cos \theta = 5 \div \sqrt{50} = \frac{1}{\sqrt{2}}$ and $\theta = 45^\circ$. Base times height is $\sqrt{5} \times \sqrt{10} \times \sin 45^\circ = \sqrt{50} \times \frac{1}{\sqrt{2}} = 5$. Both give an area of 5 square units.

The parallelogram that a = (2, 1, 0) and b = (1, 3, 0) span, seen from above. Its corners are the origin, (2, 1), (3, 4) and (1, 3), and its area is |a × b| = 5.

Sliding the top side

Slide the side b along the direction of a, to b + ka for some number k. The base and the perpendicular height do not change, so neither does the area.

The vector product agrees. a × (b + ka) = a × b + k(a × a), and a × a = 0, so a × (b + ka) = a × b.

A parallelogram with base 6 and perpendicular height 4, its top side slid 2 along the base. Drag the top corner and try to change the area: the wedge cut from one end fills the other, and the area stays 6 × 4 = 24.

The triangle is half

The diagonal from the head of a to the head of b cuts the parallelogram into two triangles. Each has sides |a| and |b| and the diagonal itself, so the two triangles are congruent, and each is half of the parallelogram.

So the triangle with sides a and b has area $\left(\frac{1}{2}\right)|a \times b| = \left(\frac{1}{2}\right)|a| |b| \sin \theta$. This is the formula $\left(\frac{1}{2}\right)ab \sin C$ for the area of a triangle, written with vectors.

If a × b has length 12, the parallelogram has area 12 and the triangle has area 12 ÷ 2 = 6.

The same parallelogram with its diagonal drawn: two congruent triangles, each half of |a × b|.

A triangle from three points

For a triangle with corners A, B and C in space, take the two sides that leave one corner as vectors. With A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3), AB = (−1, 2, 0) and AC = (−1, 0, 3).

Then AB × AC = (2 × 3 − 0 × 0, 0 × (−1) − (−1) × 3, (−1) × 0 − 2 × (−1)) = (6, 3, 2). Its length is $\sqrt{36 + 9 + 4} = \sqrt{49} = 7$, so the parallelogram on AB and AC has area 7 and the triangle ABC has area 7 ÷ 2 = 3.5 square units.

A right angle, and parallel sides

When a and b meet at a right angle, sin 90° = 1, so |a × b| = |a| |b|: the parallelogram is a rectangle, and its area is length times width. With |a| = 4 and |b| = 5 at a right angle, |a × b| = 20. This is where the vector product is largest for those lengths, and where the dot product is 0.

When a and b are parallel, $\sin \theta = 0$, the parallelogram is flat, and both the area and the vector product are 0.

The usual mistakes

Adding the components. For a × b = (2, 3, 6), the area is $\sqrt{4 + 9 + 36} = 7$, not 2 + 3 + 6 = 11.

Leaving out the square root. 49 is the area squared; the area is 7.

Giving the parallelogram when the triangle is asked for. The triangle is half of |a × b|.

Using the dot product. At a right angle a · b = 0, but the area there is as large as it can be: the area comes from the sine, through the vector product.

A shade sail

In the application below, a triangular sail is tied to the tops of three posts. The two edges from one corner are vectors, their vector product is worked out one component at a time and checked with dot products, and half of its length is the area of the sail.