01 · Length
Find |(1, −2, 2)|.
Hint
Add the squares of the three components.
Worked solution
√(1 + 4 + 4) = 3.
Understand · explore · practise
Magnitudes, unit vectors and parallelism
Calculate vector lengths, form unit vectors in a specified direction, solve magnitude conditions and test parallelism or collinearity without dividing by zero components.
Before you startThree-component vectors, displacement and Pythagoras.
01 / Length and direction answer different questions
A magnitude is a nonnegative scalar.
|v| = √(x² + y² + z²)
For v = (x, y, z), combine the three perpendicular components.
A vector includes direction. Its magnitude is only its length, written with vertical bars. A negative component does not make a length negative. Only the zero vector has magnitude zero.
Scale a vector, inspect its directionExplore
w = λv = (2, −1, 2).
|w| = |λ| |v| = 1 × 3 = 3.
Unit vector in the direction of w: (0.666667, −0.333333, 0.666667).
Blue: w. Green: its unit vector, drawn from the same origin. Both use the same projection and scale. Screen length is not true 3D length. At λ = 0 both arrows disappear because the zero vector has no direction.
02 / Square all components, then take the square root
Keep exact surds unless a decimal answer is requested.
Find the magnitude of v = (−2, 3, 6).Worked example
|v|² = (−2)² + 3² + 6² = 49
Keep brackets around negative components.
|v| = 7
Magnitude is the nonnegative square root.
02 · Exact surd
Find |(2, 1, −3)| exactly.
Hint
Do not round before taking the square root.
Worked solution
√14.
03 / Divide a nonzero vector by its magnitude
The result has length one and the same direction.
v̂ = v / |v|, provided v ≠ 0
Divide every component by the same positive scalar.
Find a unit vector in the direction of v = (2, −1, 2).Worked example
|v| = 3
The sum of squares is 9.
v̂ = (2/3, −1/3, 2/3)
Divide all components by 3.
|v̂|² = 4/9 + 1/9 + 4/9 = 1
This verifies unit length.
Watch: divide the whole vector by its length
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Another unit vector
Find a unit vector in the direction of (0, 3, −4).
Hint
The magnitude is 5.
Worked solution
(0, 3/5, −4/5).
04 / Respect the requested direction
A unit vector parallel to v can point either way.
Let v = (2, −1, 2). Compare three requests.Worked example
In the direction of v: (2/3, −1/3, 2/3)
Use the positive normalisation.
In the opposite direction: (−2/3, 1/3, −2/3)
Negate the unit vector.
Parallel to v, with no orientation specified: either of these
Parallel vectors can point in the same or opposite direction.
04 · From A towards B
A = (1, 2, 3), B = (3, 1, 5). Find a unit vector from B towards A.
Hint
Use BA rather than AB.
Worked solution
BA = (−2, 1, −2), so the required unit vector is (−2/3, 1/3, −2/3).
05 · The zero exception
Why can you not normalise (0, 0, 0)?
Hint
Its magnitude is zero.
Worked solution
Division by zero is undefined. The zero vector has no direction and therefore no unit vector in its direction.
05 / Multiply a unit vector by the requested length
Normalise first, then scale.
Find a vector of magnitude 10 in the direction of v = (0, 3, 4).Worked example
|v| = 5
Find the original length.
v / |v| = (0, 3/5, 4/5)
Normalise.
10v / |v| = (0, 6, 8)
Scale to the required length.
06 · Opposite direction and chosen length
Find a vector of magnitude 6 opposite to (2, −1, 2).
Hint
Multiply the opposite unit vector by 6.
Worked solution
(−4, 2, −4).
07 · Two possible orientations
Find all vectors of magnitude 5 parallel to (1, 2, 2).
Hint
The base magnitude is 3; both signs are allowed.
Worked solution
±(5/3, 10/3, 10/3).
06 / Use the absolute value of a scalar for length
|λv| = |λ| |v|.
v has magnitude 7 and w = −3v.Worked example
|w| = 3 × 7 = 21
Magnitude uses |−3|.
w points oppositely to v
Its negative scalar changes orientation.
A positive scale would preserve orientation
This statement requires a nonzero vector and nonzero scale.
08 · Solve a scaling condition
|v| = 4 and |λv| = 10. Find λ.
Hint
4|λ| = 10.
Worked solution
λ = 5/2 or −5/2.
07 / Magnitude equations can have two, one or no real solutions
Squaring hides the sign of an unknown component.
|(k, 2, −3)| = √29. Find k.Worked example
k² + 4 + 9 = 29
Square the magnitude equation.
k² = 16
Subtract the known contributions.
k = ±4
Both components give the same length.
09 · One solution
For |(k, 2, −3)| = √13, find k.
Hint
The known components already contribute 13.
Worked solution
k = 0.
10 · No solution
Can |(k, 2, −3)| equal 3 for real k?
Hint
Compare 9 with the fixed sum of squares 13.
Worked solution
No: k² = −4 would be required.
08 / A single scalar must match every component
For two nonzero vectors, u and v are parallel exactly when u = λv for one real scalar λ.
Test u = (2, −1, 3) and v = (−4, 2, −6).Worked example
The first component suggests λ = −2 for v = λu
Use a nonzero component.
The second and third components also multiply by −2
All three agree.
The vectors are parallel and oppositely directed
The scalar is negative.
11 · A failed test
Are (1, 2, 3) and (2, 4, 7) parallel?
Hint
The first two components suggest scale 2. Check the third.
Worked solution
No: 2 × 3 = 6, not 7.
12 · Zero components
Are (0, 2, −3) and (0, −6, 9) parallel?
Hint
Use the second component rather than forming 0/0.
Worked solution
Yes: the second vector is −3 times the first, and every component agrees.
09 / Use every component when finding a parallel vector
A zero component is a condition, not an invitation to divide by zero.
Find p, q if (p, −6, q) is parallel to (2, 3, −1).Worked example
−6 = 3λ gives λ = −2
The middle coordinate fixes the scale.
p = 2λ = −4; q = −λ = 2
Apply that same scale throughout.
(−4, −6, 2) = −2(2, 3, −1)
Substitution verifies the result.
13 · A zero condition
Find k if (1, k, 2) is parallel to (3, 0, 6).
Hint
The scale from the second vector to the first is 1/3.
Worked solution
k = 0.
10 / Points are collinear when their displacements share one line
Compare vectors drawn from a common point.
A = (1, −1, 2), B = (3, 0, 0), C = (7, 2, −4).Worked example
AB = (2, 1, −2); AC = (6, 3, −6)
Subtract the same startpoint A.
AC = 3AB
A single scalar matches all components.
A, B and C are collinear; B is between A and C
The multiplier 3 is positive and greater than 1.
14 · Location along the line
If AC = −2AB and A ≠ B, where is C relative to A and B?
Hint
The negative scale reverses direction from A.
Worked solution
C lies on the opposite side of A from B, twice as far from A as B is.
15 · Coincident points
If A = B, why is AB unsuitable as a direction for a line?
Hint
AB is the zero vector.
Worked solution
It has no direction. Use two distinct points to determine a line; three distinct points require an actual nonzero parallelism test.
11 / Check length, orientation and all components
Normalising and testing parallelism solve different problems.
- Magnitude is a nonnegative square root.
- Normalise only nonzero vectors.
- Choose the sign to match the specified direction.
- Use |λ| for the length scale.
- Retain both possible signs in magnitude equations.
- A parallelism test must agree in all three components.
- Use nonzero displacements to test a line through distinct points.
16 · One calculation, two conclusions
u = (−2, 1, −2), v = (6, −3, 6). State their relationship and a unit vector in the direction of v.
Hint
Compare v with u, then normalise v.
Worked solution
v = −3u, so they are parallel and oppositely directed. |v| = 9; v/|v| = (2/3, −1/3, 2/3).
Section 1 of 11 · Length and direction answer different questions