Magnitude of a vector
Section: Transformations and vectors | Syllabus: Cambridge IGCSE Mathematics (0580)
Finding the Magnitude of a Vector
Extended Only
The magnitude of a vector is its length. It is also called the modulus of the vector, and is always a positive value - the direction of the vector does not affect it.
- The magnitude of AB is written |AB|; the magnitude of a is written |a|
- In component form, the horizontal and vertical components of a vector form the two shorter sides of a right-angled triangle, and the magnitude is the hypotenuse
- For a vector x. y , apply Pythagoras' theorem: |a|=√(x^2+y^2)
Worked Example: Finding the Magnitude of a Vector
- Question: Find the magnitude of the vector a= 8. -6 .
- Step 1: Apply the formula: |a|=√(8^2+(-6)^2)
- Step 2: =√(64+36)=√(100)=10
- Answer: |a|=10
Worked Example: A Magnitude That Does Not Simplify Exactly
- Question: Find the magnitude of the vector b= 7. -3 , correct to 3 significant figures.
- Step 1: |b|=√(7^2+(-3)^2)=√(49+9)=√(58)
- Step 2: Since 58 is not a perfect square, evaluate on a calculator: √(58)7.6158...
- Answer: |b|7.62 (3 s.f.)
Common Mistakes
MistakeGiving a negative magnitude when one or both components of the vector are negative
Fixsquaring a negative component always gives a positive result, so a magnitude is always positive, regardless of the vector's direction
MistakeAdding the components before squaring, e.g. working out (x+y)² instead of x²+y²
Fixsquare each component separately first, then add the two squared values together, exactly as in Pythagoras' theorem
Using the Magnitude to Find an Unknown Component
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Some problems give the magnitude of a vector and ask you to work backwards to find a missing coordinate or component. Since a magnitude is always positive, this usually produces two possible values - one from each square root.
- Set up the magnitude equation x^2+y^2=|a|^2 using the known and unknown components
- Rearrange to isolate the unknown squared term, then take the square root of both sides
- Remember both the positive and negative square root - each gives a valid solution unless the context rules one out
Worked Example: Finding an Unknown Coordinate
- Question: Point P has coordinates (-1,4) and point Q has coordinates (k,4). Given that |PQ|=3√(5), find the two possible values of k.
- Step 1: Find PQ: (k-(-1),\,4-4)=(k+1,\,0)
- Step 2: Square the magnitude: (k+1)^2+0^2=(3√(5))^2=45
- Step 3: k+1=±√(45)=3√(5)
- Answer: k=3√(5)-1 or k=-3√(5)-1
Common Mistakes
MistakeOnly giving one value of the unknown, forgetting that a squared equation has two roots
Fixwhenever you take a square root to solve for an unknown component, write out both the + and - solutions before finishing
MistakeSquaring the given magnitude incorrectly when it is written as a surd, e.g. treating (3√(5))^2 as 3√(5)2
Fix(a√(b))^2=a^2× b - square the whole number and the surd separately, then multiply
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