5.1.3 Vector Magnitude: The Measure

Discover how vector length is just Pythagoras in disguise. Modulus notation: |a|

Mode A: Triangle Scanner

Magnitude: 5.00 units

x-component: 3

y-component: 4

√(x² + y²) = √(3² + 4²)

= √(9 + 16)

= √25

= 5.00 units

The magnitude |a| is just the hypotenuse of the right-angled triangle formed by the x and y components.

|a| = √(x² + y²)

This is in the Edexcel formula sheet (Appendix 5)

The hypotenuse length is 5.00, always longer than either component alone.

Negative components: When you square them, they become positive! So |⟨-5, 12⟩| = √(25 + 144) = 13, not √(-25 + 144).

Length is never negative. The modulus is always ≥ 0.

x-component: -5

y-component: 12

Question: Find the magnitude of vectorAB = ⟨-5, 12⟩

STEP 1

Write the formula:

|AB| = √(x² + y²)

STEP 2

Substitute x = -5, y = 12:

|AB| = √((-5)² + (12)²)

STEP 3

Square each component:

|AB| = √(25 + 144)

✓ Negative squared = positive

STEP 4

Add the squares:

|AB| = √169

ANSWER

Calculate the square root:

|AB| = 13.00 units

✓ Perfect square! Exact answer: 13

Mark Scheme: You must show the substitution and squaring steps. Just writing √169 = 13.00 loses marks for missing working (AO1).

Edexcel 4MA1 Higher Tier | Syllabus 5.1 E & F | Formula Sheet: Appendix 5

You've reached the end of the free preview. Sign in — and upgrade to Plus or Pro — to keep going with Math4MA1.

Full Access

Unlock all 158 lessons