Area of a Triangle Notes

Published on: September 29, 2026

This Area of a Triangle Calculator finds the area of the triangle two vectors outline, and shows the working line by line. The cross product gives the area of the whole parallelogram those vectors span, and the triangle is exactly half of it. Halving at the end is the only difference between the two results. Following the same pattern every time keeps the halving from being applied too early.

Step-by-step method

  1. Set up the problem.
  2. Compute the cross product A × B.
  3. Use Area = 1/2 · |A × B| and simplify.
See Example 1 Hide Example 1

Problem

\(\vec{A} = \left\langle 1,\; 0,\; 0 \right\rangle,\; \vec{B} = \left\langle 0,\; 1,\; 0 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the cross product A × B.
  3. Use Area = 1/2 · |A × B| and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 1,\; 0,\; 0 \right\rangle\) and \(\vec{B} = \left\langle 0,\; 1,\; 0 \right\rangle\).

\(\vec{A} = \left\langle 1,\; 0,\; 0 \right\rangle,\; \vec{B} = \left\langle 0,\; 1,\; 0 \right\rangle\)

Step 2

Step 2 - Compute the cross product A × B.

In this problem: The cross product is \(\left\langle 0,\; 0,\; 1 \right\rangle\).

\(\vec{A} \times \vec{B} = \left\langle 00 - 01,\; 00 - 10,\; 11 - 00 \right\rangle = \left\langle 0,\; 0,\; 1 \right\rangle\)

Step 3

Step 3 - Use Area = 1/2 · |A × B| and simplify.

In this problem: The triangle area is half the parallelogram area.

\(\text{Area} = \frac{1}{2}\sqrt{0^{2} + 0^{2} + 1^{2}} = \frac{1}{2} \cdot 1 = \frac{1}{2}\)

Final Answer

\(\text{Area} = \frac{1}{2}\)
See Example 2 Hide Example 2

Problem

\(\vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle,\; \vec{B} = \left\langle 4,\; 5,\; 6 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the cross product A × B.
  3. Use Area = 1/2 · |A × B| and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle\) and \(\vec{B} = \left\langle 4,\; 5,\; 6 \right\rangle\).

\(\vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle,\; \vec{B} = \left\langle 4,\; 5,\; 6 \right\rangle\)

Step 2

Step 2 - Compute the cross product A × B.

In this problem: The cross product is \(\left\langle -3,\; 6,\; -3 \right\rangle\).

\(\vec{A} \times \vec{B} = \left\langle 26 - 35,\; 34 - 16,\; 15 - 24 \right\rangle = \left\langle -3,\; 6,\; -3 \right\rangle\)

Step 3

Step 3 - Use Area = 1/2 · |A × B| and simplify.

In this problem: The triangle area is half the parallelogram area.

\(\text{Area} = \frac{1}{2}\sqrt{\left(-3\right)^{2} + 6^{2} + \left(-3\right)^{2}} = \frac{1}{2} \cdot 3 \sqrt{6} = \frac{3 \sqrt{6}}{2} \approx 3.67\)

Final Answer

\(\text{Area} = \frac{3 \sqrt{6}}{2} \approx 3.67\)
See Example 3 Hide Example 3

Problem

\(\vec{A} = \left\langle 2,\; 0,\; 1 \right\rangle,\; \vec{B} = \left\langle 1,\; 3,\; 0 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the cross product A × B.
  3. Use Area = 1/2 · |A × B| and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 2,\; 0,\; 1 \right\rangle\) and \(\vec{B} = \left\langle 1,\; 3,\; 0 \right\rangle\).

\(\vec{A} = \left\langle 2,\; 0,\; 1 \right\rangle,\; \vec{B} = \left\langle 1,\; 3,\; 0 \right\rangle\)

Step 2

Step 2 - Compute the cross product A × B.

In this problem: The cross product is \(\left\langle -3,\; 1,\; 6 \right\rangle\).

\(\vec{A} \times \vec{B} = \left\langle 00 - 13,\; 11 - 20,\; 23 - 01 \right\rangle = \left\langle -3,\; 1,\; 6 \right\rangle\)

Step 3

Step 3 - Use Area = 1/2 · |A × B| and simplify.

In this problem: The triangle area is half the parallelogram area.

\(\text{Area} = \frac{1}{2}\sqrt{\left(-3\right)^{2} + 1^{2} + 6^{2}} = \frac{1}{2} \cdot \sqrt{46} = \frac{\sqrt{46}}{2} \approx 3.39\)

Final Answer

\(\text{Area} = \frac{\sqrt{46}}{2} \approx 3.39\)
See Example 4 Hide Example 4

Problem

\(\vec{A} = \left\langle 3,\; 1,\; -2 \right\rangle,\; \vec{B} = \left\langle 1,\; 2,\; 4 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the cross product A × B.
  3. Use Area = 1/2 · |A × B| and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 3,\; 1,\; -2 \right\rangle\) and \(\vec{B} = \left\langle 1,\; 2,\; 4 \right\rangle\).

\(\vec{A} = \left\langle 3,\; 1,\; -2 \right\rangle,\; \vec{B} = \left\langle 1,\; 2,\; 4 \right\rangle\)

Step 2

Step 2 - Compute the cross product A × B.

In this problem: The cross product is \(\left\langle 8,\; -14,\; 5 \right\rangle\).

\(\vec{A} \times \vec{B} = \left\langle 14 - \left(-2\right)2,\; \left(-2\right)1 - 34,\; 32 - 11 \right\rangle = \left\langle 8,\; -14,\; 5 \right\rangle\)

Step 3

Step 3 - Use Area = 1/2 · |A × B| and simplify.

In this problem: The triangle area is half the parallelogram area.

\(\text{Area} = \frac{1}{2}\sqrt{8^{2} + \left(-14\right)^{2} + 5^{2}} = \frac{1}{2} \cdot \sqrt{285} = \frac{\sqrt{285}}{2} \approx 8.44\)

Final Answer

\(\text{Area} = \frac{\sqrt{285}}{2} \approx 8.44\)