01 · Coefficient comparison
For independent a, b, c, (p + 1)a + 2b + qc = 4a + rb − 3c. Find p, q, r.
Hint
Compare each independent coefficient.
Worked solution
p = 3, q = −3, r = 2.
Understand · explore · practise
Vector proofs in three dimensions
Build vector proofs using independent directions, compare every component, verify line intersections and prove diagonal and midpoint results in three-dimensional shapes.
Before you startVector arithmetic, position vectors, ratios, noncollinear points and simultaneous equations.
01 / A proof must establish the relationship for every allowed shape
Choose an origin and name the displacement vectors clearly.
OP = xa + yb + zc
The coefficients describe how the chosen vectors combine to reach P.
Coordinate calculations can prove a claim for specified points. A general vector proof uses arbitrary vectors subject to the geometric assumptions. A diagram or one numerical example can suggest a result, but does not by itself prove it for all shapes.
Agreement in two coordinates is not enoughExplore
A(s) = (s, 2s, 0); B(t) = (1 − t, t, h).
A(s) = (1/3, 2/3, 0); B(t) = (1/3, 2/3, 0).
All three coordinates agree: the selected points coincide.
Amber and blue mark points on their respective lines. The drawing is a projection. Check the full coordinate readout to establish whether the points really coincide in space.
02 / Compare coefficients only in independent directions
Three noncoplanar vectors give independent directions in space.
xa + yb + zc = x′a + y′b + z′c ⇒ x = x′, y = y′, z = z′
This implication requires a, b and c to be linearly independent.
Let a, b, c be noncoplanar. Solve 2a + pb − c = qa + 3b + rc.Worked example
Compare the coefficient of a: q = 2
No combination of the other independent directions can replace a.
Compare b: p = 3
The representation is unique.
Compare c: r = −1
All three coefficients must agree.
03 / Pairwise nonparallel does not mean three-way independence
Three directions can all lie in one plane.
a = (1, 0, 0), b = (0, 1, 0), c = (1, 1, 0).Worked example
No pair is parallel
None is a scalar multiple of another.
c = a + b
The third vector is a combination of the first two.
a + b − c = 0
Nonzero coefficients produce the zero vector, so coefficient comparison is not justified.
02 · Two representations
Give two coefficient triples for the vector c in this example.
Hint
Use c itself, or a + b.
Worked solution
c = 0a + 0b + 1c = 1a + 1b + 0c.
03 · State the missing assumption
A proof compares coefficients of three pairwise nonparallel vectors. What must it establish first?
Hint
Think about a third vector lying in the plane of the first two.
Worked solution
It needs independence, for example that the three vectors are noncoplanar.
04 / Represent a point on a line by a starting point and a direction
r = a + s(b − a) is a position-vector description.
Line A(s) has points (s, 2s, 0). Line B(t) has points (1 − t, t, 0). Find an intersection.Worked example
s = 1 − t; 2s = t; 0 = 0
Equate all coordinates.
3s = 1, so s = 1/3 and t = 2/3
Solve the nontrivial equations.
Both points equal (1/3, 2/3, 0)
Substitution into both lines completes the check.
Watch: check the third coordinate too
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 · Different parameters
Why need s and t not be equal at an intersection?
Hint
The lines have different starting points and direction vectors.
Worked solution
They are independent line parameters. An intersection requires equal position vectors, not equal parameter values.
05 / A candidate from two equations must satisfy the third
Two projections can meet while the actual lines do not.
Replace B(t) by (1 − t, t, 1).Worked example
The first two equations still give s = 1/3, t = 2/3
The horizontal coordinates can agree.
The third equation would require 0 = 1
It is impossible.
There is no intersection
Checking only two coordinates would give a false conclusion.
05 · A vertical gap
For A(s) = (s, 2s, 0) and B(t) = (1 − t, t, h), which h permits an intersection?
Hint
Use the third coordinate.
Worked solution
h = 0 only.
06 · Complete a candidate check
A candidate gives positions (2, −1, 4) and (2, −1, 5). Is it an intersection?
Hint
All coordinates must agree.
Worked solution
No. Their z coordinates differ.
06 / Prove that parallelogram diagonals bisect each other
Use arbitrary adjacent side vectors.
Take O, A, D, C in parallelogram boundary order, with OA = a, OD = b, OC = a + b.Worked example
Midpoint of OC has position (a + b)/2
Average the two endpoint positions.
Midpoint of AD has position (a + b)/2
The other endpoints are a and b.
The diagonal midpoints coincide
This proves the result for every nondegenerate parallelogram.
07 · A three-dimensional parallelepiped
A parallelepiped has adjacent vectors a, b, c from O. Find the midpoint of its space diagonal from O to a + b + c.
Hint
Average the endpoint positions.
Worked solution
(a + b + c)/2.
08 · Another space diagonal
Show that the segment from a to b + c has the same midpoint.
Hint
Add its endpoint positions and divide by 2.
Worked solution
[a + (b + c)]/2 = (a + b + c)/2.
07 / A triangle centroid lies two thirds along each median
Derive its position from a vertex and an opposite midpoint.
Triangle ABC has position vectors a, b, c. Let M be the midpoint of BC.Worked example
m = (b + c)/2
First locate the opposite midpoint.
g = a + (2/3)(m − a)
Travel two thirds from A towards M.
g = (a + b + c)/3
The expression is symmetric in the three vertices.
The other two medians reach the same point
So all three medians meet there.
09 · Centroid coordinates
Find the centroid of (1, 0, 2), (4, 3, −1), (−2, 3, 5).
Hint
Average each coordinate across the three vertices.
Worked solution
(1, 2, 2).
08 / Locate two triangle centroids on a space diagonal
The coefficients reveal the fractions along the diagonal.
In a parallelepiped, three adjacent vertices have positions a, b, c; the opposite vertex is e = a + b + c.Worked example
Their triangle centroid g₁ = (a + b + c)/3 = e/3
It lies one third of the way from O to E.
The opposite triangle has vertices a + b, b + c, c + a
Add these three position vectors.
Its centroid g₂ = 2(a + b + c)/3 = 2e/3
It lies two thirds of the way from O to E.
O, G₁, G₂, E divide the diagonal into three equal parts
Each successive displacement is e/3.
10 · Diagonal length fractions
If the space diagonal has length 12, what are OG₁ and OG₂?
Hint
Use the positive scalar multiples 1/3 and 2/3.
Worked solution
4 and 8 respectively.
11 · Directed differences
Find G₁G₂ in terms of e.
Hint
Subtract the starting position from the ending position.
Worked solution
2e/3 − e/3 = e/3.
09 / Opposite-edge midpoint segments in a tetrahedron concur
Find a shared midpoint algebraically.
Tetrahedron OABC has OA = a, OB = b, OC = c.Worked example
Midpoints of OA and BC are a/2 and (b + c)/2
These are opposite edges.
The midpoint of the joining segment is (a + b + c)/4
Average their position vectors.
Pair OB with AC, or OC with AB
Both constructions have exactly the same midpoint.
The three joining segments meet at that common point
The calculation is symmetric in a, b and c.
12 · A numerical tetrahedron
O = (0, 0, 0), A = (4, 0, 0), B = (0, 8, 0), C = (0, 0, 12). Find this common point.
Hint
Use (a + b + c)/4.
Worked solution
(1, 2, 3).
13 · Opposite face centroid
The centroid F of face ABC is (a + b + c)/3. What fraction of OF locates the common point G?
Hint
Compare g with f.
Worked solution
g = (3/4)f, so OG : GF = 3 : 1.
10 / State what each algebraic result proves
A scalar multiple proves parallelism; a point formula needs a starting point.
Suppose you obtain PQ = (1/2)AB.Worked example
If AB ≠ 0, PQ is parallel to AB and half its length
The scalar is positive, so orientation agrees.
This does not by itself prove the two segments are on the same line
Their starting points may differ.
To prove P lies on AB, show p = a + t(b − a)
This fixes both position and direction.
14 · Parallel but not collinear
Give two equal displacement vectors whose segments lie on different lines.
Hint
Translate a segment without changing its displacement.
Worked solution
(0,0,0) → (1,0,0) and (0,1,0) → (1,1,0) both have displacement (1,0,0), but lie on different parallel lines.
15 · Example versus proof
Does checking a midpoint identity for one cube prove it for every parallelepiped?
Hint
A cube is a special case.
Worked solution
No. A proof with arbitrary independent adjacent vectors covers the general shape.
11 / Choose assumptions, derive positions, then verify the conclusion
A concise proof still needs all its conditions.
- Choose and state an origin.
- Use arbitrary vectors for a general proof.
- Establish independence before comparing basis coefficients.
- Use separate parameters for separate lines.
- Check every component of an intersection.
- Use position averages for midpoint and centroid results.
- Distinguish a parallel direction from the location of a line.
16 · Final coefficient check
For noncoplanar a, b, c, solve sa + 2sb + c = a + tb + rc.
Hint
Match all three independent coefficients.
Worked solution
s = 1, t = 2 and r = 1.
Section 1 of 11 · A proof must establish the relationship for every allowed shape