Magnitude of a Vector Calculator

Published on: February 22, 2026
Final Answer: Free • Full Steps: Pro

This Magnitude of a Vector Calculator measures how long a vector is in three dimensions, and shows the working line by line. Each component is squared, the squares are added, and the square root of that total is the length, which is Pythagoras applied twice. Squaring is what removes the signs, so direction never affects the answer. Following the same pattern every time keeps the length positive and easy to check.

Step-by-step method

  1. Set up the problem.
  2. Write the magnitude formula.
  3. Substitute and compute inside the square root.
  4. Simplify the square root (if possible).
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. Write the magnitude formula.
  3. Substitute and compute inside the square root.
  4. Simplify the square root (if possible).

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 - Write the magnitude formula.

In this problem: The magnitude is the square root of the sum of the squared components.

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

Step 3

Step 3 - Substitute and compute inside the square root.

In this problem: The value inside the square root is \(25\).

\(\left|\vec{A}\right| = \sqrt{3^{2} + 4^{2} + 0^{2}} = \sqrt{9 + 16 + 0} = \sqrt{25}\)

Step 4

Step 4 - Simplify the square root (if possible).

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

\(\left|\vec{A}\right| = 5\)

Final Answer

\(\left|\vec{A}\right| = 5\)
See Example 2 Hide Example 2

Problem

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

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the magnitude formula.
  3. Substitute and compute inside the square root.
  4. Simplify the square root (if possible).

Step 1

Step 1 - Set up the problem.

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

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

Step 2

Step 2 - Write the magnitude formula.

In this problem: The magnitude is the square root of the sum of the squared components.

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

Step 3

Step 3 - Substitute and compute inside the square root.

In this problem: The value inside the square root is \(14\).

\(\left|\vec{A}\right| = \sqrt{1^{2} + 2^{2} + 3^{2}} = \sqrt{1 + 4 + 9} = \sqrt{14}\)

Step 4

Step 4 - Simplify the square root (if possible).

In this problem: The magnitude is \(\sqrt{14}\).

\(\left|\vec{A}\right| = \sqrt{14} \approx 3.74\)

Final Answer

\(\left|\vec{A}\right| = \sqrt{14} \approx 3.74\)

Example Problems

Click one to load it into the box and solve it.