Area of a Parallelogram Notes

Published on: September 29, 2026

This Area of a Parallelogram Calculator finds the area spanned by two vectors, and shows the working line by line. The cross product is built first, because its length is exactly the area of the parallelogram the two vectors outline. Vectors pointing the same way span nothing, which is why a zero cross product means zero area. Following the same pattern every time keeps the cross product and its magnitude as separate moves.

Step-by-step method

  1. Set up the problem.
  2. Compute the cross product A × B.
  3. Compute the magnitude |A × B| to get the area.
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. Compute the magnitude |A × B| to get the area.

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 - Compute the magnitude |A × B| to get the area.

In this problem: The area equals the magnitude of the cross product.

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

Final Answer

\(\text{Area} = 1\)
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. Compute the magnitude |A × B| to get the area.

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 - Compute the magnitude |A × B| to get the area.

In this problem: The area equals the magnitude of the cross product.

\(\text{Area} = \sqrt{\left(-3\right)^{2} + 6^{2} + \left(-3\right)^{2}} = 3 \sqrt{6} \approx 7.35\)

Final Answer

\(\text{Area} = 3 \sqrt{6} \approx 7.35\)
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. Compute the magnitude |A × B| to get the area.

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 - Compute the magnitude |A × B| to get the area.

In this problem: The area equals the magnitude of the cross product.

\(\text{Area} = \sqrt{\left(-3\right)^{2} + 1^{2} + 6^{2}} = \sqrt{46} \approx 6.78\)

Final Answer

\(\text{Area} = \sqrt{46} \approx 6.78\)
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. Compute the magnitude |A × B| to get the area.

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 - Compute the magnitude |A × B| to get the area.

In this problem: The area equals the magnitude of the cross product.

\(\text{Area} = \sqrt{8^{2} + \left(-14\right)^{2} + 5^{2}} = \sqrt{285} \approx 16.88\)

Final Answer

\(\text{Area} = \sqrt{285} \approx 16.88\)