Vector Magnitude and Direction Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Vector Magnitude and Direction.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

The magnitude (or length) of a vector \mathbf{v} = \langle v_1, v_2 \rangle is \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}, calculated using the Pythagorean theorem. A unit vector has magnitude 1 and indicates direction only. The unit vector in the direction of \mathbf{v} is \hat{\mathbf{v}} = \frac{\mathbf{v}}{\|\mathbf{v}\|}.

Magnitude is how long the arrow is—like measuring the length of a stick. Direction is which way it points. A unit vector is a 'pure direction' with length 1, like a compass needle. To get the unit vector, shrink or stretch the vector until its length is exactly 1 while keeping it pointed the same way.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Every vector can be decomposed into magnitude (how much) and direction (which way). Unit vectors encode pure direction.

Common stuck point: Do not forget to take the square root when computing magnitude. Also, \arctan alone does not always give the correct angle—check the quadrant of the vector.

Sense of Study hint: When you need magnitude, draw the vector as a right triangle with components as legs, then apply the Pythagorean theorem: \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}. For the unit vector, divide each component by the magnitude. For direction, use \theta = \arctan(v_2/v_1) and adjust for the correct quadrant.

Worked Examples

Example 1

easy
Find the magnitude of \mathbf{v} = \langle 3, 4 \rangle.

Solution

  1. 1
    Step 1: \|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25}.
  2. 2
    Step 2: = 5.
  3. 3
    Check: This is a 3-4-5 right triangle ✓

Answer

5
The magnitude (length) of a vector is found using the Pythagorean theorem: \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}. This extends naturally to higher dimensions.

Example 2

medium
Find the unit vector in the direction of \mathbf{v} = \langle 1, 2, 2 \rangle.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Find \|\langle -5, 12 \rangle\|.

Example 2

hard
Find the direction angle \theta of \mathbf{v} = \langle -1, \sqrt{3} \rangle measured from the positive x-axis.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

vector operationssimplifying radicals