Area of Triangles and Parallelograms Using Vectors

This lesson finds areas using the vector cross product. For two adjacent side vectors sharing the same starting point, the area of a triangle is half the magnitude of their cross product, and the area of a parallelogram (or rectangle) is the full magnitude.

Area of a triangle

Find the area of the triangle with vertices $A(1,1,2)$, $B(2,3,5)$ and $C(1,5,5)$.

Form two adjacent sides from $A$:

$\vec{AB} = (2-1)\,\hat{\imath} + (3-1)\,\hat{\jmath} + (5-2)\,\hat{k} = \hat{\imath} + 2\hat{\jmath} + 3\hat{k}$

$\vec{AC} = (1-1)\,\hat{\imath} + (5-1)\,\hat{\jmath} + (5-2)\,\hat{k} = 0\,\hat{\imath} + 4\hat{\jmath} + 3\hat{k}$

The area is $\tfrac{1}{2}\,\lvert \vec{AB} \times \vec{AC} \rvert$. Compute the cross product as a determinant:

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix}$

Expanding along the first row:

$= \hat{\imath}(2\cdot 3 - 3\cdot 4) - \hat{\jmath}(1\cdot 3 - 3\cdot 0) + \hat{k}(1\cdot 4 - 2\cdot 0) = -6\,\hat{\imath} - 3\hat{\jmath} + 4\hat{k}$

Then the magnitude is

$\lvert \vec{AB} \times \vec{AC} \rvert = \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$

So the area of triangle $ABC$ is

$\text{Area} = \tfrac{1}{2}\sqrt{61} \text{ square units.}$

Area of a parallelogram

Find the area of the parallelogram whose adjacent sides are $\vec{a} = \hat{\imath} - \hat{\jmath} + 3\hat{k}$ and $\vec{b} = 2\hat{\imath} - 7\hat{\jmath} + \hat{k}$. The area is $\lvert \vec{a} \times \vec{b} \rvert$.

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix}$

Expanding:

$= \hat{\imath}\big((-1)(1) - (3)(-7)\big) - \hat{\jmath}\big((1)(1) - (3)(2)\big) + \hat{k}\big((1)(-7) - (-1)(2)\big)$

$= \hat{\imath}(-1 + 21) - \hat{\jmath}(1 - 6) + \hat{k}(-7 + 2) = 20\,\hat{\imath} + 5\hat{\jmath} - 5\hat{k}$

Its magnitude is

$\lvert \vec{a} \times \vec{b} \rvert = \sqrt{20^2 + 5^2 + (-5)^2} = \sqrt{400 + 25 + 25} = \sqrt{450} = \sqrt{225 \cdot 2} = 15\sqrt{2}$

So the area is $15\sqrt{2}$ square units.

Area of a rectangle

Find the area of the rectangle with vertices $A$, $B$, $C$, $D$ whose position vectors are

$A = -\hat{\imath} + \tfrac{1}{2}\hat{\jmath} + 4\hat{k}, \quad B = \hat{\imath} + \tfrac{1}{2}\hat{\jmath} + 4\hat{k}$

$C = \hat{\imath} - \tfrac{1}{2}\hat{\jmath} + 4\hat{k}, \quad D = -\hat{\imath} - \tfrac{1}{2}\hat{\jmath} + 4\hat{k}$

A rectangle is a parallelogram, so its area is the magnitude of the cross product of two adjacent sides. Build them by subtracting position vectors.

Side AB

$\vec{AB} = B - A = \big(1 - (-1)\big)\hat{\imath} + \big(\tfrac{1}{2} - \tfrac{1}{2}\big)\hat{\jmath} + (4 - 4)\hat{k} = 2\hat{\imath} + 0\hat{\jmath} + 0\hat{k}$

Side AD

$\vec{AD} = D - A = \big(-1 - (-1)\big)\hat{\imath} + \big(-\tfrac{1}{2} - \tfrac{1}{2}\big)\hat{\jmath} + (4 - 4)\hat{k} = 0\hat{\imath} - \hat{\jmath} + 0\hat{k}$

Now the cross product:

$\vec{AB} \times \vec{AD} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 0 & 0 \\ 0 & -1 & 0 \end{vmatrix} = \hat{k}\big((2)(-1) - (0)(0)\big) = -2\hat{k}$

Its magnitude is

$\lvert \vec{AB} \times \vec{AD} \rvert = \sqrt{(-2)^2} = \sqrt{4} = 2$

So the area of the rectangle is $2$ square units.

Key takeaways

• The cross product of two adjacent side vectors gives a vector whose length equals the area of the parallelogram they span.
• Area of a triangle is $\tfrac{1}{2}\,\lvert \vec{AB} \times \vec{AC} \rvert$; area of a parallelogram or rectangle is $\lvert \vec{AB} \times \vec{AD} \rvert$.
• Build the side vectors by subtracting position vectors, then evaluate the cross product as a $3\times 3$ determinant.