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How to Calculate Area of a Triangle with Vectors

Finding the area of a triangle spanned by $\vec{u}$ and $\vec{v}$ is equal to the length of the $\vec{u} \times \vec{v}$ -vector divided by two. That means it is half the area of a parallelogram. If you know the angle between the two vectors, you can use this formula:

Formula

The Area of a Triangle with a Known Angle

\begin{array}{l} \frac{1}{2} \cdot | \vec{u} \times \vec{v} | = \frac{1}{2} \cdot | \vec{u} | \cdot | \vec{v} | \cdot \sin α, \\ α = ∠ (\vec{u}, \vec{v}) \end{array}

\frac{1}{2} \cdot | \vec{u} \times \vec{v} | = \frac{1}{2} \cdot | \vec{u} | \cdot | \vec{v} | \cdot \sin α, α = ∠ (\vec{u}, \vec{v})

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

Formula

The Area of a Triangle on Vector Coordinate Form

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

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

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

Example 1

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

First, you have to find the cross product of the vectors, which turns out to be $(16, 2, 11)$ . The length of this vector will be equal to the area of the parallelogram $\vec{u}$ and $\vec{v}$ spans. That means you have to divide the length by $2$ to find the area of the triangle.

\begin{array}{l} \frac{1}{2} \sqrt{1 6^{2} + 2^{2} + 1 1^{2}} & = \frac{1}{2} \sqrt{256 + 4 + 121} \\ = \frac{1}{2} \sqrt{381} \\ \approx 9.8 \end{array}

The area of the triangle is approximately equal to $9.8$ .

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