Angle Between Vectors Notes

Published on: September 29, 2026

This Angle Between Vectors Calculator finds the angle separating two vectors in three dimensions, and shows the working line by line. The dot product divided by the two lengths gives the cosine of that angle, because the dot product already measures how much the vectors agree. Taking the inverse cosine turns that ratio back into an angle. Following the same pattern every time keeps the lengths and the dot product in their right places.

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product A · B.
  3. Compute the magnitudes |A| and |B|.
  4. Use θ = arccos((A · B) / (|A||B|)) and evaluate.
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 dot product A · B.
  3. Compute the magnitudes |A| and |B|.
  4. Use θ = arccos((A · B) / (|A||B|)) and evaluate.

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 dot product A · B.

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

\(\vec{A} \cdot \vec{B} = 10 + 01 + 00 = 0\)

Step 3

Step 3 - Compute the magnitudes |A| and |B|.

In this problem: The magnitudes are \(1\) and \(1\).

\(\left|\vec{A}\right| = 1,\quad \left|\vec{B}\right| = 1\)

Step 4

Step 4 - Use θ = arccos((A · B) / (|A||B|)) and evaluate.

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

\(\theta = \arccos\left(\frac{0}{1 \cdot 1}\right) \approx 90^{\circ}\)

Final Answer

\(\theta \approx 90^{\circ}\)
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 dot product A · B.
  3. Compute the magnitudes |A| and |B|.
  4. Use θ = arccos((A · B) / (|A||B|)) and evaluate.

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 dot product A · B.

In this problem: The dot product is \(32\).

\(\vec{A} \cdot \vec{B} = 14 + 25 + 36 = 32\)

Step 3

Step 3 - Compute the magnitudes |A| and |B|.

In this problem: The magnitudes are \(\sqrt{14}\) and \(\sqrt{77}\).

\(\left|\vec{A}\right| = \sqrt{14},\quad \left|\vec{B}\right| = \sqrt{77}\)

Step 4

Step 4 - Use θ = arccos((A · B) / (|A||B|)) and evaluate.

In this problem: The angle is approximately \(12.9332^{\circ}\).

\(\theta = \arccos\left(\frac{32}{\sqrt{14} \cdot \sqrt{77}}\right) \approx 12.9332^{\circ}\)

Final Answer

\(\theta \approx 12.9332^{\circ}\)
See Example 3 Hide Example 3

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product A · B.
  3. Compute the magnitudes |A| and |B|.
  4. Use θ = arccos((A · B) / (|A||B|)) and evaluate.

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Compute the dot product A · B.

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

\(\vec{A} \cdot \vec{B} = 21 + 10 + 21 = 4\)

Step 3

Step 3 - Compute the magnitudes |A| and |B|.

In this problem: The magnitudes are \(3\) and \(\sqrt{2}\).

\(\left|\vec{A}\right| = 3,\quad \left|\vec{B}\right| = \sqrt{2}\)

Step 4

Step 4 - Use θ = arccos((A · B) / (|A||B|)) and evaluate.

In this problem: The angle is approximately \(19.4712^{\circ}\).

\(\theta = \arccos\left(\frac{4}{3 \cdot \sqrt{2}}\right) \approx 19.4712^{\circ}\)

Final Answer

\(\theta \approx 19.4712^{\circ}\)
See Example 4 Hide Example 4

Problem

\(\vec{A} = \left\langle 3,\; -1,\; 2 \right\rangle,\; \vec{B} = \left\langle 1,\; 4,\; 0 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product A · B.
  3. Compute the magnitudes |A| and |B|.
  4. Use θ = arccos((A · B) / (|A||B|)) and evaluate.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 3,\; -1,\; 2 \right\rangle\) and \(\vec{B} = \left\langle 1,\; 4,\; 0 \right\rangle\).

\(\vec{A} = \left\langle 3,\; -1,\; 2 \right\rangle,\; \vec{B} = \left\langle 1,\; 4,\; 0 \right\rangle\)

Step 2

Step 2 - Compute the dot product A · B.

In this problem: The dot product is \(-1\).

\(\vec{A} \cdot \vec{B} = 31 + \left(-1\right)4 + 20 = -1\)

Step 3

Step 3 - Compute the magnitudes |A| and |B|.

In this problem: The magnitudes are \(\sqrt{14}\) and \(\sqrt{17}\).

\(\left|\vec{A}\right| = \sqrt{14},\quad \left|\vec{B}\right| = \sqrt{17}\)

Step 4

Step 4 - Use θ = arccos((A · B) / (|A||B|)) and evaluate.

In this problem: The angle is approximately \(93.7165^{\circ}\).

\(\theta = \arccos\left(\frac{-1}{\sqrt{14} \cdot \sqrt{17}}\right) \approx 93.7165^{\circ}\)

Final Answer

\(\theta \approx 93.7165^{\circ}\)