Angle Between Two Planes (3D) Notes

Published on: September 29, 2026

This Angle Between Two Planes (3D) Calculator finds the angle at which two planes meet, and shows the working line by line. The angle between planes is the same as the angle between their normal vectors, so the coefficients of each equation are read off as a vector first. The dot product then gives the cosine of that angle. Following the same pattern every time keeps the normals from being confused with the planes themselves.

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product |n₁ · n₂|.
  3. Compute the magnitudes |n₁| and |n₂|.
  4. Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.
See Example 1 Hide Example 1

Problem

\(2x - 3y + 4z = 10,\quad x + 4y - 2z = 7\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product |n₁ · n₂|.
  3. Compute the magnitudes |n₁| and |n₂|.
  4. Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Write the angle formula.

In this problem: The angle between planes is the angle between their normal vectors.

\(\cos\left(\theta\right) = \frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\)

Step 3

Step 3 - Compute the dot product |n₁ · n₂|.

In this problem: The absolute dot product is \(18\).

\(\left|\vec{n}_{1} \cdot \vec{n}_{2}\right| = 18\)

Step 4

Step 4 - Compute the magnitudes |n₁| and |n₂|.

In this problem: The magnitudes are \(\sqrt{29}\) and \(\sqrt{21}\).

\(\left|\vec{n}_{1}\right| = \sqrt{29},\quad \left|\vec{n}_{2}\right| = \sqrt{21}\)

Step 5

Step 5 - Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.

In this problem: The angle between the planes is approximately \(43.1641^{\circ}\).

\(\theta = \arccos\left(\frac{18}{\sqrt{29} \cdot \sqrt{21}}\right) \approx 43.1641^{\circ}\)

Final Answer

\(\theta \approx 43.1641^{\circ}\)
See Example 2 Hide Example 2

Problem

\(x + y + z = 6,\quad 2x - y + 2z = 7\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product |n₁ · n₂|.
  3. Compute the magnitudes |n₁| and |n₂|.
  4. Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.

Step 1

Step 1 - Set up the problem.

In this problem: The normals are \(\vec{n}_{1} = \left\langle 1,\; 1,\; 1 \right\rangle\) and \(\vec{n}_{2} = \left\langle 2,\; -1,\; 2 \right\rangle\).

\(\vec{n}_{1} = \left\langle 1,\; 1,\; 1 \right\rangle,\; \vec{n}_{2} = \left\langle 2,\; -1,\; 2 \right\rangle\)

Step 2

Step 2 - Write the angle formula.

In this problem: The angle between planes is the angle between their normal vectors.

\(\cos\left(\theta\right) = \frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\)

Step 3

Step 3 - Compute the dot product |n₁ · n₂|.

In this problem: The absolute dot product is \(3\).

\(\left|\vec{n}_{1} \cdot \vec{n}_{2}\right| = 3\)

Step 4

Step 4 - Compute the magnitudes |n₁| and |n₂|.

In this problem: The magnitudes are \(\sqrt{3}\) and \(3\).

\(\left|\vec{n}_{1}\right| = \sqrt{3},\quad \left|\vec{n}_{2}\right| = 3\)

Step 5

Step 5 - Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.

In this problem: The angle between the planes is approximately \(54.7356^{\circ}\).

\(\theta = \arccos\left(\frac{3}{\sqrt{3} \cdot 3}\right) \approx 54.7356^{\circ}\)

Final Answer

\(\theta \approx 54.7356^{\circ}\)
See Example 3 Hide Example 3

Problem

\(x + 2y + 2z = 4,\quad 2x + y - 2z = 5\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product |n₁ · n₂|.
  3. Compute the magnitudes |n₁| and |n₂|.
  4. Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.

Step 1

Step 1 - Set up the problem.

In this problem: The normals are \(\vec{n}_{1} = \left\langle 1,\; 2,\; 2 \right\rangle\) and \(\vec{n}_{2} = \left\langle 2,\; 1,\; -2 \right\rangle\).

\(\vec{n}_{1} = \left\langle 1,\; 2,\; 2 \right\rangle,\; \vec{n}_{2} = \left\langle 2,\; 1,\; -2 \right\rangle\)

Step 2

Step 2 - Write the angle formula.

In this problem: The angle between planes is the angle between their normal vectors.

\(\cos\left(\theta\right) = \frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\)

Step 3

Step 3 - Compute the dot product |n₁ · n₂|.

In this problem: The absolute dot product is \(0\).

\(\left|\vec{n}_{1} \cdot \vec{n}_{2}\right| = 0\)

Step 4

Step 4 - Compute the magnitudes |n₁| and |n₂|.

In this problem: The magnitudes are \(3\) and \(3\).

\(\left|\vec{n}_{1}\right| = 3,\quad \left|\vec{n}_{2}\right| = 3\)

Step 5

Step 5 - Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.

In this problem: The angle between the planes is approximately \(90^{\circ}\).

\(\theta = \arccos\left(\frac{0}{3 \cdot 3}\right) \approx 90^{\circ}\)

Final Answer

\(\theta \approx 90^{\circ}\)
See Example 4 Hide Example 4

Problem

\(3x - y + 2z = 8,\quad x + y + z = 3\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product |n₁ · n₂|.
  3. Compute the magnitudes |n₁| and |n₂|.
  4. Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.

Step 1

Step 1 - Set up the problem.

In this problem: The normals are \(\vec{n}_{1} = \left\langle 3,\; -1,\; 2 \right\rangle\) and \(\vec{n}_{2} = \left\langle 1,\; 1,\; 1 \right\rangle\).

\(\vec{n}_{1} = \left\langle 3,\; -1,\; 2 \right\rangle,\; \vec{n}_{2} = \left\langle 1,\; 1,\; 1 \right\rangle\)

Step 2

Step 2 - Write the angle formula.

In this problem: The angle between planes is the angle between their normal vectors.

\(\cos\left(\theta\right) = \frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\)

Step 3

Step 3 - Compute the dot product |n₁ · n₂|.

In this problem: The absolute dot product is \(4\).

\(\left|\vec{n}_{1} \cdot \vec{n}_{2}\right| = 4\)

Step 4

Step 4 - Compute the magnitudes |n₁| and |n₂|.

In this problem: The magnitudes are \(\sqrt{14}\) and \(\sqrt{3}\).

\(\left|\vec{n}_{1}\right| = \sqrt{14},\quad \left|\vec{n}_{2}\right| = \sqrt{3}\)

Step 5

Step 5 - Use cos(θ) = |n₁ · n₂| / (|n₁||n₂|) and solve for θ.

In this problem: The angle between the planes is approximately \(51.8871^{\circ}\).

\(\theta = \arccos\left(\frac{4}{\sqrt{14} \cdot \sqrt{3}}\right) \approx 51.8871^{\circ}\)

Final Answer

\(\theta \approx 51.8871^{\circ}\)