Vector geometry.
Use routes, position vectors and scalar multiples to explain geometric relationships.
Find position vectors and ratios · show parallelism and collinearity · reason with similarity and intersections
A route is the sum of its directed steps
Vector geometry describes a diagram through routes. To travel from A to B, you can move directly or follow several connected arrows; the total displacement is the same if the start and finish match.
A position vector begins at an agreed origin O. Subtracting two position vectors gives a vector between their endpoints. Draw the route A → O → B to see why AB equals OB minus OA, rather than trying to memorise a subtraction order without a picture.
Vectors add along a route. Reversing a step changes its sign. A position vector starts at a chosen origin O.
Pause and explainWhy is the vector from A back to O the negative of OA?
It has the same length but the reverse direction, so every component changes sign.
Use a ratio to find a position vector
A ratio along a segment tells you what fraction of the entire segment has been travelled. If AM : MB is 2 : 3, the complete segment contains five equal parts and AM contains two of them.
Find AM as that fraction of AB, then add OA to obtain the position vector OM. This two-route view explains the weighting formula. It also prevents confusing the ratio of the two pieces with the fraction of the whole segment.
OA = a, OB = b, and M divides AB internally in the ratio AM : MB = 2 : 3. Find OM.
M is 25 of the way from A to B, not 23 of the way.
OM = a + 25(b − a)
= 35a + 25b
In general, if AM : MB = m : n, then OM = nm + na + mm + nb. The weights on the endpoints sum to 1.
Pause and explainIf AM : MB = 2 : 3, why is AM two fifths of AB rather than two thirds?
AB includes both pieces, so it has 2 + 3 = 5 ratio parts. AM uses 2 of those 5.
Scalar multiples prove parallelism
Vectors that are non-zero scalar multiples point along parallel lines. Scaling changes length and possibly direction, but does not turn the arrow away from its line.
Express both candidate vectors in terms of the same two non-parallel basis vectors. Factor one expression to reveal a scalar multiple of the other. When a triangle is cut by a parallel segment, that same factor can establish a length ratio and then a squared area ratio.
M and N are midpoints of OA and OB. Show MN is parallel to AB.
MN = 12b − 12a = 12(b − a) = 12AB
MN is a non-zero scalar multiple of AB, so the lines are parallel. MN has half the length of AB. Triangles OMN and OAB are similar, with length ratio 1 : 2 and area ratio 1 : 4.
Pause and explainIf MN is half of AB as a vector, what can you conclude?
The segments are parallel, and MN has half the length of AB, assuming AB is non-zero.
Prove collinearity and compare coefficients
Parallel direction alone does not put three points on one line. For collinearity, compare vectors that share a point, such as PQ and PR, and show that one is a scalar multiple of the other.
The shared point anchors the parallel directions to one actual line. When comparing coefficients of a and b, the vectors a and b must be non-parallel; otherwise different coefficient combinations could describe the same direction.
OP = 2a + b, OQ = 5a + 4b, OR = 8a + 7b. Show P, Q and R are collinear.
PR = 6a + 6b = 2PQ
PQ and PR have the common starting point P and are parallel. Therefore P, Q and R are collinear; Q is the midpoint of PR.
OP = a + 2b, OQ = 3a + kb. Find k if O, P and Q are collinear.
Write OQ = λOP. Comparing the a coefficients gives λ = 3. Comparing the b coefficients gives k = 2λ = 6.
Pause and explainWhy are parallel vectors on two unrelated parts of a diagram not enough to prove collinearity?
They can lie on different parallel lines. A common point is needed to establish one shared line.
Two routes to the same point
An intersection belongs to two routes at the same time. Describing its position in two ways creates an equation between vector expressions, which can determine where the routes meet.
Assign a fraction parameter to each route, expand the two expressions, then compare coefficients of the same non-parallel basis vectors. Interpret the resulting fractions geometrically: one half means a midpoint; another fraction gives a division ratio.
OABC is a parallelogram, with OA = a and OC = b. Show its diagonals OB and AC bisect each other.
OB = a + b. If M is on OB, write OM = λ(a + b).
Using the route along AC, OM = a + μ(b − a) = (1 − μ)a + μb.
λ = μ = 12
Both parameters are half, so M is the midpoint of both diagonals.
Pause and explainIf an intersection is halfway along each diagonal of a parallelogram, what property has been shown?
The diagonals bisect each other: each cuts the other into two equal segments.
Your turn. Give a vector reason.
Draw and label the route you use before writing algebra. State which vectors are position vectors, which are reversed, and what fraction of a segment is involved.
Finish each proof with its geometric conclusion. A scalar multiple should lead explicitly to parallelism or collinearity, with the common point stated when needed.
OA = a and OB = b. Express BA.
Reveal worked answer
M divides AB internally with AM : MB = 2 : 1. Express OM in terms of OA = a and OB = b.
Reveal worked answer
OP = 2a − b and OQ = 6a − 3b. Explain why O, P and Q are collinear.
Reveal worked answer
OM = 34OA and ON = 34OB. Express MN in terms of AB, and give the area ratio of triangles OMN and OAB.
Reveal worked answer
In parallelogram OABC with OA = a and OC = b, give OB and the position vector of its midpoint.