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 $\theta$θ between a and b, that height is the side opposite $\theta$θ, $|b|\sin \theta$|b| sin θ.

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

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

The length of the vector product

The length of a × b is exactly that area: $|a\times b|=|a||b|\sin \theta$|a × b| = |a| |b| sin θ. 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 $a=\left(a_1,a_2,0\right)$a = (a₁, a₂, 0) and $b=\left(b_1,b_2,0\right)$b = (b₁, b₂, 0). Their vector product is $\left(0,0,a_1{b}_2-a_2{b}_1\right)$(0, 0, a₁b₂ − a₂b₁), so $|a\times b{|}^2={\left(a_1{b}_2-a_2{b}_1\right)}^2$|a × b|² = (a₁b₂ − a₂b₁)². Now work out $|a{|}^2|b{|}^2-{\left(a\cdot b\right)}^2=\left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)-{\left(a_1{b}_1+a_2{b}_2\right)}^2$|a|² |b|² − (a · b)² = (a₁² + a₂²)(b₁² + b₂²) − (a₁b₁ + a₂b₂)². The terms $a_1^2{b}_1^2$a₁²b₁² and $a_2^2{b}_2^2$a₂²b₂² cancel, leaving $a_1^2{b}_2^2+a_2^2{b}_1^2-2a_1{a}_2{b}_1{b}_2$a₁²b₂² + a₂²b₁² − 2a₁a₂b₁b₂, which is ${\left(a_1{b}_2-a_2{b}_1\right)}^2$(a₁b₂ − a₂b₁)². So $|a\times b{|}^2=|a{|}^2|b{|}^2-{\left(a\cdot b\right)}^2$|a × b|² = |a|² |b|² − (a · b)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 · b = |a| |b| cos θ: $|a\times b{|}^2=|a{|}^2|b{|}^2-|a{|}^2|b{|}^2{\cos }^2\theta =|a{|}^2|b{|}^2\left(1-{\cos }^2\theta \right)$|a × b|² = |a|² |b|² − |a|² |b|² cos²θ = |a|² |b|² (1 − cos²θ). Since ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1, that is $|a{|}^2|b{|}^2{\sin }^2\theta$|a|² |b|² sin²θ. For $\theta$θ from 0° to 180°, $\sin \theta$sin θ is not negative, so $|a\times b|=|a||b|\sin \theta$|a × b| = |a| |b| sin θ.

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: $|a|=\sqrt{5}$|a| = √5, $|b|=\sqrt{10}$|b| = √10 and a · b = 2 + 3 = 5, so $\cos \theta =5÷\sqrt{50}=\frac{1}{\sqrt{2}}$cos θ = 5 ÷ √50 = 1/√2 and $\theta =45^{\circ }$θ = 45°. Base times height is $\sqrt{5}\times \sqrt{10}\times \sin 45^{\circ }=\sqrt{50}\times \frac{1}{\sqrt{2}}=5$√5 × √10 × sin 45° = √50 × 1/√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 $\left(\frac{1}{2}\right)|a\times b|=\left(\frac{1}{2}\right)|a||b|\sin \theta$(1/2)|a × b| = (1/2)|a| |b| sin θ. This is the formula $\left(\frac{1}{2}\right)ab\sin C$(1/2)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 $\sqrt{36+9+4}=\sqrt{49}=7$√(36 + 9 + 4) = √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, $\sin \theta =0$sin θ = 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 $\sqrt{4+9+36}=7$√(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.