For v = xi + yj, the magnitude is √(x² + y²), and the direction is an angle from the positive x-axis found with tan⁻¹ and a sketch of the quadrant.
This lesson is part of vectors. It assumes you can already write vectors in component and unit-vector form.
How do I find magnitude and direction?
Follow the same steps each time.
- Magnitude: |v| = √(x² + y²).
- Reference angle: α = tan⁻¹(|y| ÷ |x|), always acute.
- Sketch the vector to see its quadrant.
- Direction: quadrant 1, θ = α. Quadrant 2, θ = 180° − α. Quadrant 3, θ = 180° + α. Quadrant 4, θ = 360° − α.
Worked example 1: a vector in the second quadrant
Find the magnitude and direction of v = −5i + 12j.
|v| = √((−5)² + 12²) = √(25 + 144) = √169 = 13.
The reference angle is α = tan⁻¹(12 ÷ 5) = 67.38°. The vector has negative x and positive y, so it lies in quadrant 2, and θ = 180° − 67.38° = 112.6°.
Worked example 2: a unit vector and a vector from polar data
The unit vector along a = 3i + 4j is a ÷ |a| = (3i + 4j) ÷ 5 = 0.6i + 0.8j. Check its length: √(0.36 + 0.64) = 1.
Now build a vector of magnitude 10 at 30° above the x-axis: x = 10 cos 30° = 5√3 ≈ 8.66 and y = 10 sin 30° = 5, so the vector is 8.66i + 5j. Its magnitude is √(75 + 25) = 10.
Worked example 3: magnitude of a sum
Let a = 3i + 4j and b = −i + 2j. Find |a + b|.
a + b = 2i + 6j, so |a + b| = √(4 + 36) = √40 = 2√10 ≈ 6.32. Compare |a| + |b| = 5 + √5 ≈ 7.24, which is larger. The magnitude of a sum is not the sum of magnitudes.
The mistake that costs marks
The common slip is to quote the calculator’s tan⁻¹(y ÷ x) directly. For v = −5i + 12j, tan⁻¹(12 ÷ (−5)) = −67.38°, which points into quadrant 4, not quadrant 2.
| Step | Wrong | Right |
|---|---|---|
| Calculator value | tan⁻¹(12 ÷ −5) = −67.38° | α = tan⁻¹(12 ÷ 5) = 67.38° |
| Quadrant | (not checked) | Negative x, positive y: quadrant 2 |
| Direction | −67.38° | 180° − 67.38° = 112.6° |
A sketch takes five seconds and decides the quadrant. Make it the third step of every direction question.
Check yourself
Find the magnitude and direction of v = 6i − 6j.
Answer
|v| = √(36 + 36) = √72 = 6√2 ≈ 8.49.
The reference angle is tan⁻¹(6 ÷ 6) = 45°. The vector has positive x and negative y, so it lies in quadrant 4: θ = 360° − 45° = 315°, which can also be written as −45°.
What to study next
Test the whole chapter with the vectors practice set. The next skill is using parallel vectors and collinearity, which uses the vectors you can now build and measure.
If you want a teacher to work through magnitude and direction with you, see online one-to-one Additional Mathematics tuition.