Vector Magnitude and Direction Math Example 3

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Example 3

medium
Find the magnitude and direction angle of the vector v = <3, 4>.

Solution

  1. 1
    Find the magnitude: ∥v∥=32+42=9+16=25=5\|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.
  2. 2
    Find the direction angle θ\theta using tan⁡θ=yx=43\tan\theta = \frac{y}{x} = \frac{4}{3}.
  3. 3
    Compute θ=arctan⁡(43)≈53.13°\theta = \arctan\left(\frac{4}{3}\right) \approx 53.13°. Since both components are positive, the vector is in Quadrant I, so no adjustment is needed.

Answer

Magnitude =5= 5, direction angle ≈53.13°\approx 53.13°
The magnitude uses the Pythagorean theorem. The direction angle is measured counterclockwise from the positive xx-axis. This is a 3-4-5 right triangle, one of the most common Pythagorean triples.

About Vector Magnitude and Direction

The magnitude ∥v∥\|\mathbf{v}\| is a vector's length; the direction is the angle it makes with a reference axis.

Learn more about Vector Magnitude and Direction →

More Vector Magnitude and Direction Examples

Example 1 easy

Find the magnitude of [formula].

Example 2 medium

Find the unit vector in the direction of [formula].

Example 4 easy

Find [formula].

Example 5 hard

Find the direction angle [formula] of [formula] measured from the positive [formula]-axis.

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