Unit Vector Notes

Published on: September 29, 2026

This Unit Vector Calculator rescales a vector so that it has length one while pointing the same way, and shows the working line by line. The length is measured first, and then every component is divided by it, which is what forces the new length to be exactly one. The result carries only the direction and none of the size. Following the same pattern every time keeps the direction unchanged while the length is normalised.

Step-by-step method

  1. Set up the problem.
  2. Compute the magnitude |A|.
  3. Use  = A / |A| and simplify.
See Example 1 Hide Example 1

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the magnitude |A|.
  3. Use  = A / |A| and simplify.

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Compute the magnitude |A|.

In this problem: The magnitude is \(5\).

\(\left|\vec{A}\right| = \sqrt{3^{2} + 4^{2} + 0^{2}} = 5\)

Step 3

Step 3 - Use  = A / |A| and simplify.

In this problem: Divide every component by the magnitude.

\(\hat{A} = \frac{1}{5}\left\langle 3,\; 4,\; 0 \right\rangle = \left\langle \frac{3}{5},\; \frac{4}{5},\; 0 \right\rangle\)

Final Answer

\(\hat{A} = \left\langle \frac{3}{5},\; \frac{4}{5},\; 0 \right\rangle\)
See Example 2 Hide Example 2

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the magnitude |A|.
  3. Use  = A / |A| and simplify.

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Compute the magnitude |A|.

In this problem: The magnitude is \(3\).

\(\left|\vec{A}\right| = \sqrt{1^{2} + 2^{2} + 2^{2}} = 3\)

Step 3

Step 3 - Use  = A / |A| and simplify.

In this problem: Divide every component by the magnitude.

\(\hat{A} = \frac{1}{3}\left\langle 1,\; 2,\; 2 \right\rangle = \left\langle \frac{1}{3},\; \frac{2}{3},\; \frac{2}{3} \right\rangle\)

Final Answer

\(\hat{A} = \left\langle \frac{1}{3},\; \frac{2}{3},\; \frac{2}{3} \right\rangle\)
See Example 3 Hide Example 3

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the magnitude |A|.
  3. Use  = A / |A| and simplify.

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Compute the magnitude |A|.

In this problem: The magnitude is \(7\).

\(\left|\vec{A}\right| = \sqrt{2^{2} + 3^{2} + 6^{2}} = 7\)

Step 3

Step 3 - Use  = A / |A| and simplify.

In this problem: Divide every component by the magnitude.

\(\hat{A} = \frac{1}{7}\left\langle 2,\; 3,\; 6 \right\rangle = \left\langle \frac{2}{7},\; \frac{3}{7},\; \frac{6}{7} \right\rangle\)

Final Answer

\(\hat{A} = \left\langle \frac{2}{7},\; \frac{3}{7},\; \frac{6}{7} \right\rangle\)
See Example 4 Hide Example 4

Problem

\(\vec{A} = \left\langle 0,\; 5,\; 12 \right\rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Compute the magnitude |A|.
  3. Use  = A / |A| and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The components are \(\vec{A} = \left\langle 0,\; 5,\; 12 \right\rangle\).

\(\vec{A} = \left\langle 0,\; 5,\; 12 \right\rangle\)

Step 2

Step 2 - Compute the magnitude |A|.

In this problem: The magnitude is \(13\).

\(\left|\vec{A}\right| = \sqrt{0^{2} + 5^{2} + 12^{2}} = 13\)

Step 3

Step 3 - Use  = A / |A| and simplify.

In this problem: Divide every component by the magnitude.

\(\hat{A} = \frac{1}{13}\left\langle 0,\; 5,\; 12 \right\rangle = \left\langle 0,\; \frac{5}{13},\; \frac{12}{13} \right\rangle\)

Final Answer

\(\hat{A} = \left\langle 0,\; \frac{5}{13},\; \frac{12}{13} \right\rangle\)