Projection of A onto B Notes

Published on: September 29, 2026

This Projection of A onto B Calculator finds the shadow one vector casts along another, and shows the working line by line. The dot product measures how much of the first vector lies along the second, and dividing by the second vector's squared length rescales that into a multiple of it. What is left points along the second vector only. Following the same pattern every time keeps the two vectors from being swapped.

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product A · B.
  3. Compute |B|².
  4. Use proj_B(A) = ((A · B) / |B|²) B and simplify.
See Example 1 Hide Example 1

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 |B|².
  4. Use proj_B(A) = ((A · B) / |B|²) 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 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 |B|².

In this problem: The squared magnitude of B is \(77\).

\(\left|\vec{B}\right|^{2} = 4^{2} + 5^{2} + 6^{2} = 77\)

Step 4

Step 4 - Use proj_B(A) = ((A · B) / |B|²) B and simplify.

In this problem: Scale \(\vec{B}\) by \(\frac{32}{77}\).

\(\operatorname{proj}_{\vec{B}} \vec{A} = \frac{32}{77} \left\langle 4,\; 5,\; 6 \right\rangle = \left\langle \frac{128}{77},\; \frac{160}{77},\; \frac{192}{77} \right\rangle\)

Final Answer

\(\operatorname{proj}_{\vec{B}} \vec{A} = \left\langle \frac{128}{77},\; \frac{160}{77},\; \frac{192}{77} \right\rangle\)
See Example 2 Hide Example 2

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product A · B.
  3. Compute |B|².
  4. Use proj_B(A) = ((A · B) / |B|²) B and simplify.

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Compute the dot product A · B.

In this problem: The dot product is \(2\).

\(\vec{A} \cdot \vec{B} = 21 + 01 + 00 = 2\)

Step 3

Step 3 - Compute |B|².

In this problem: The squared magnitude of B is \(2\).

\(\left|\vec{B}\right|^{2} = 1^{2} + 1^{2} + 0^{2} = 2\)

Step 4

Step 4 - Use proj_B(A) = ((A · B) / |B|²) B and simplify.

In this problem: Scale \(\vec{B}\) by \(1\).

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

Final Answer

\(\operatorname{proj}_{\vec{B}} \vec{A} = \left\langle 1,\; 1,\; 0 \right\rangle\)
See Example 3 Hide Example 3

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product A · B.
  3. Compute |B|².
  4. Use proj_B(A) = ((A · B) / |B|²) B and simplify.

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Compute the dot product A · B.

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

\(\vec{A} \cdot \vec{B} = 31 + 40 + 00 = 3\)

Step 3

Step 3 - Compute |B|².

In this problem: The squared magnitude of B is \(1\).

\(\left|\vec{B}\right|^{2} = 1^{2} + 0^{2} + 0^{2} = 1\)

Step 4

Step 4 - Use proj_B(A) = ((A · B) / |B|²) B and simplify.

In this problem: Scale \(\vec{B}\) by \(3\).

\(\operatorname{proj}_{\vec{B}} \vec{A} = \frac{3}{1} \left\langle 1,\; 0,\; 0 \right\rangle = \left\langle 3,\; 0,\; 0 \right\rangle\)

Final Answer

\(\operatorname{proj}_{\vec{B}} \vec{A} = \left\langle 3,\; 0,\; 0 \right\rangle\)
See Example 4 Hide Example 4

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the dot product A · B.
  3. Compute |B|².
  4. Use proj_B(A) = ((A · B) / |B|²) B and simplify.

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Compute the dot product A · B.

In this problem: The dot product is \(6\).

\(\vec{A} \cdot \vec{B} = 21 + \left(-1\right)2 + 32 = 6\)

Step 3

Step 3 - Compute |B|².

In this problem: The squared magnitude of B is \(9\).

\(\left|\vec{B}\right|^{2} = 1^{2} + 2^{2} + 2^{2} = 9\)

Step 4

Step 4 - Use proj_B(A) = ((A · B) / |B|²) B and simplify.

In this problem: Scale \(\vec{B}\) by \(\frac{2}{3}\).

\(\operatorname{proj}_{\vec{B}} \vec{A} = \frac{6}{9} \left\langle 1,\; 2,\; 2 \right\rangle = \left\langle \frac{2}{3},\; \frac{4}{3},\; \frac{4}{3} \right\rangle\)

Final Answer

\(\operatorname{proj}_{\vec{B}} \vec{A} = \left\langle \frac{2}{3},\; \frac{4}{3},\; \frac{4}{3} \right\rangle\)