How to Calculate Area of a Parallelogram with Vectors

Finding the area of a parallelogram spanned by $\vec{u}$ and $\vec{v}$ is the same as calculating the length of the $\vec{u} \times \vec{v}$ -vector.

Two vectors and the area of the parallelogram spanned by the vectors

If you know the angle between the two vectors, you can use this formula:

Formula

| \vec{u} \times \vec{v} | = | \vec{u} | \cdot | \vec{v} | \cdot \sin α, α = ∠ (\vec{u}, \vec{v})

If you have the vectors on vector coordinate form, you use this formula:

Formula

\begin{array}{l} | \vec{u} \times \vec{v} | & = | \begin{matrix} (x_{1}, & y_{1}, & z_{1}) \\ \times \\ (x_{2}, & y_{2}, & z_{2}) \end{matrix} | \\ = | (y_{1} z_{2} - y_{2} z_{1}, z_{1} x_{2} - z_{2} x_{1}, \\ x_{1} y_{2} - x_{2} y_{1}) | \end{array}

| \vec{u} \times \vec{v} | = | \begin{matrix} (x_{1}, & y_{1}, & z_{1}) \\ \times \\ (x_{2}, & y_{2}, & z_{2}) \end{matrix} | = | (y_{1} z_{2} - y_{2} z_{1}, z_{1} x_{2} - z_{2} x_{1}, x_{1} y_{2} - x_{2} y_{1}) |

Example 1

Find the area of the parallelogram that is spanned by $\vec{u} = (1, 3, - 2)$ and $\vec{v} = (- 3, 2, 4)$ .

You start by finding the cross product:

\begin{array}{l} \vec{u} \times \vec{v} = \begin{matrix} (1, & 3, & - 2) \\ \times \\ (- 3, & 2, & 4) \end{matrix} \\ = (3 \cdot 4 - 2 \cdot (- 2), (- 2) \cdot (- 3) - 1 \cdot 4, \\ 1 \cdot 2 - (- 3) \cdot 3) \\ = (12 + 4, 6 - 4, 2 + 9) \\ = (16, 2, 11) \end{array}

\begin{array}{l} \vec{u} \times \vec{v} & = \begin{matrix} (1, & 3, & - 2) \\ \times \\ (- 3, & 2, & 4) \end{matrix} \\ = (3 \cdot 4 - 2 \cdot (- 2), (- 2) \cdot (- 3) - 1 \cdot 4, 1 \cdot 2 - (- 3) \cdot 3) \\ = (12 + 4, 6 - 4, 2 + 9) \\ = (16, 2, 11) \end{array}

The length of this vector will now be the area of the parallelogram. You find the length like this:

\begin{array}{l} \sqrt{1 6^{2} + 2^{2} + 1 1^{2}} & = \sqrt{256 + 4 + 121} \\ = \sqrt{381} \\ \approx 19.5 \end{array}

\sqrt{1 6^{2} + 2^{2} + 1 1^{2}} = \sqrt{256 + 4 + 121} = \sqrt{381} \approx 19.5

The area of the parallelogram is approximately equal to

19.5

Example 2

If you have two vectors

\vec{a}

and

\vec{b}

, where

| \vec{a} | = 5

| \vec{b} | = 7

, and the angle between them is

30

°, you can find the area of the parallelogram they span by inserting your information into the formula:

\begin{array}{l} | \vec{a} \times \vec{b} | & = | \vec{a} | \cdot | \vec{b} | \cdot \sin α \\ = 5 \cdot 7 \cdot \sin 30 ° \\ = 5 \cdot 7 \cdot \frac{1}{2} \\ = \frac{35}{2} \\ = 17.5 . \end{array}

The area of the parallelogram is equal to $17.5$ .

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