0580 · 7.4Vector geometry

Vector properties and scalar multiples

0:00A geometry diagram can hide relationships that are difficult to see. Two lines might look parallel, but appearance alone cannot prove it. Vectors let us replace that uncertainty with exact reasoning. By describing journeys between labelled positions, we can prove parallel lines, identify collinear points, and calculate ratios without measuring the diagram. The key idea is that different routes can produce the same overall movement. Going directly from P to Q has the same result as travelling through O.

0:28We will use that connection throughout this video. Every expression will describe a journey, rather than just letters to rearrange. You need a little vector addition, fraction arithmetic, and algebraic simplification. Vector addition is covered in the video on vectors in two dimensions. Here, we will use those skills to reason about geometry. This section is Extended content, and the methods must work without a calculator. A vector has both magnitude and direction.

0:55We represent it by a directed line segment, which is drawn with an arrowhead. The length represents its magnitude, while the arrow shows its direction. Vector PQ means the movement from P to Q, not the reverse journey. Equal vectors have the same magnitude and direction, even when they start at different points. Moving an arrow without turning or stretching it leaves its vector unchanged. Reversing the arrow gives the negative of the original vector.

1:19In examination questions, vector names are printed in bold. Multiplying a vector by a positive number changes its magnitude but preserves its direction. Multiplication by a negative number also reverses the direction. For example, negative two times vector p points opposite to vector p and has twice its magnitude.

Position vectors and routes

1:39This explains why non-zero scalar multiples give parallel directions. A position vector describes a point relative to a chosen origin. If O is our origin, vector p is the position vector of P. Vector q similarly locates Q. These vectors start at the same reference point, so the difference tells us how to travel between the two endpoints. To travel from P to Q through O, first reverse vector p.

2:05That first movement is negative vector p, taking us from P back to O. Then follow vector q from O to Q. Adding those movements gives vector q minus vector p. This gives the rule behind position vector subtraction. A vector between two points equals the destination position vector minus the starting position vector. The order matters because swapping the two vectors reverses the journey. Vector QP is therefore vector p minus vector q.

2:32We can also add several vectors along a connected route. Going from P to R and then from R to Q gives vector PQ. The intermediate point disappears from the overall journey. A route returning to its starting point has a total of the zero vector, because it ends where it began. Our examples use two coplanar vectors, meaning both lie in the same plane. We will choose vector p and vector q in different directions.

2:57Their combinations can describe journeys across the diagram. Keeping these two basic directions lets us compare expressions without introducing a new vector for every segment.

Parallel lines from ratios

3:07Our first worked example asks us to express vector MN in terms of vector p and vector q. Point M is the midpoint of OP, and N lies halfway along OQ. We must also show that MN is parallel to PQ. Begin by translating the midpoint information into position vectors. The position vector of M is one half of vector p because M lies halfway from O to P. Likewise, vector ON is one half of vector q.

3:36Subtracting the starting position from the destination gives vector MN as one half vector q − ½ vector p. Factorising gives one half multiplied by the difference between vector q and vector p. Vector PQ is vector q minus vector p. Substituting that result gives vector MN equal to one half of vector PQ. The vectors are non-zero scalar multiples, so their directions are parallel. This is the reason for the conclusion, rather than simply evidence from the drawing.

4:05The multiplier also tells us that MN is half the length of PQ. Triangle OMN is similar to triangle OPQ. The shared angle at O and corresponding angles on parallel lines establish the similarity. All corresponding lengths therefore have the same ratio, linking our vector result to familiar geometry. Parallel vectors need not be equal vectors. In this example, the directions match, but the magnitudes differ. A negative multiplier still establishes parallel directions, although the arrows then point opposite ways.

4:38The multiplier must apply to the whole vector expression, not just one term. The same method handles a ratio other than one half. Suppose G divides OP in the ratio two to one, and H divides OQ in the same ratio. Each side then has three equal parts, so vector OG is two thirds of vector p. Likewise, vector OH is two thirds of vector q. Vector GH equals vector OH minus vector OG, which is two thirds vector q − ⅔ vector p.

5:09Factorising gives two thirds of vector PQ. So GH is parallel to PQ, and its length is two thirds of the length of PQ. Triangle OGH is similar to triangle OPQ, with corresponding lengths in the ratio two to three.

Proving points are collinear

5:24Our second example concerns collinear points, which lie on one straight line. The position vectors of P and Q are vector p and vector q. The position vector of R is three vector q minus vector p, all divided by two. We must show that P, Q and R are collinear and find the ratio PQ to QR. First, rewrite vector OR as 3/2 vector q − ½ vector p.

5:50To obtain vector PR, subtract vector p from that position vector. Combining the vector p terms gives 3/2 vector q − 3/2 vector p. Factorising produces 3/2 multiplied by the difference between vector q and vector p. Vector PQ equals vector q minus vector p. Therefore, vector PR is 3/2 times vector PQ. These parallel vectors share the starting point P, so their endpoints lie on the same straight line.

6:19Consequently, P, Q and R are collinear. The shared point is essential to this argument. Parallel arrows elsewhere in a diagram do not establish that three chosen points lie on one line. Here, both journeys begin at P, which fixes one common line. The positive multiplier greater than one also places R beyond Q. To find vector QR, subtract vector PQ from vector PR. 3/2 − 1 leaves one half of vector PQ.

6:46The length ratio PQ to QR is therefore one to one half. Multiplying both parts by two gives the ratio two to one. Ratios describe parts of a segment, so always identify the whole carefully. If D divides OP in the ratio OD to DP of one to two, the whole has three equal parts. Vector OD is consequently one third of vector p. Using one half would compare the first part with the second, rather than with the whole.

Intersecting lines and coefficients

7:14Our third example combines position vectors, ratios and intersecting lines. In triangle OPQ, D divides OP in the ratio one to two. Point E divides PQ in the ratio two to one. The lines OE and QD meet at X, and we must find OX to XE. The ratio on OP gives vector OD equal to one third vector p. Along PQ, vector PE is two thirds of the difference between vector q and vector p.

7:44Travelling from O through P gives vector OE as vector p plus that expression. Expanding and collecting terms produces one third vector p + ⅔ vector q. Point X lies on two different lines, so we can describe its position by two routes. Let t be the fraction of the journey from O to E that reaches X. Use the letter s for the fraction of the journey from Q to D that reaches X.

8:08Both descriptions must give the same position vector. Using OE, vector OX is t times the bracket one third vector p + ⅔ vector q. Expanding gives t/3 vector p plus two t over three vector q. The other route starts with vector q and adds s times vector QD, which is one third vector p minus vector q. Collecting terms gives s/3 vector p, plus the bracket 1 − s multiplied by vector q.

8:37The two expressions describe exactly the same position. Because vector p and vector q are not parallel, their coefficients must match separately. Extra movement along vector p can never make up for a shortfall along vector q. We therefore compare the vector p coefficients first, then those of vector q. Matching the vector p coefficients gives t/3 equal to s/3, so t = s. The vector q coefficients give 2t/3 equal to 1 − s.

9:06Substituting t for s gives 2t/3 equal to 1 − t. Adding t gives 5t/3 equal to one, so t is three fifths. This means OX occupies three fifths of OE. The remaining segment XE occupies two fifths of the whole. Therefore, OX to XE is three fifths to two fifths. Simplifying gives the required ratio three to two. The value of t is not itself the final ratio. It compares OX with the whole segment OE.

9:36To compare OX with XE, we must include the remaining fraction. This distinction prevents a common incorrect answer of three to five.

Exam method and recap

9:45In an examination, watch the command words express, show and find. Express a vector by choosing a connected route and substituting the given vectors. Show parallel vectors by writing one as a scalar multiple of the other. Show collinear points by adding the fact that the parallel journeys share a point. Several common errors come from losing the meaning of the arrows. Reversing the subtraction describes the opposite journey. Assuming parallelism from the picture replaces proof with guesswork.

10:14Matching coefficients works only when the two given vectors are not parallel, and different names alone are not enough. These methods must work on Paper Two without a calculator, and also on Paper Four, where a calculator is allowed. Keep fractional coefficients exact throughout the working. These examples require no rounding. If a question later requests a decimal, follow its accuracy instruction and round only the final answer. Vector geometry turns a diagram into connected journeys.

10:40Position vectors locate points from an origin, and their differences describe movement from one point to another. Scalar multiples establish parallel directions, while a shared point turns that relationship into a collinearity proof. Ratios tell us how far along a journey each point lies. Whenever a diagram becomes complicated, return to the starting point and destination of the required vector. Choose a route whose pieces you already know. If two routes reach the same point, their vector expressions must agree.

11:10That simple principle is what makes these geometric proofs work.