E7.4 Vector geometry

Learning objectives

E7.4.1Represent vectors by directed line segments.• Represent vectors by directed line segments.E7.4.2Use position vectors.• Use position vectors.E7.4.3Use the sum and difference of two or more vectors to express given• Use the sum and difference of two or more vectors to express given vectors in terms of two coplanar vectors.E7.4.4Use vectors to reason and to solve geometric problems• Use vectors to reason and to solve geometric problems. Examples include: • show that vectors are parallel • show that 3 points are collinear • solve vector problems involving ratio and similarity.

A directed line segment represents both a movement and its direction. In $\overrightarrow{AB}$ , $A$ is the tail and $B$ is the head; the arrow points from the starting point to the ending point.

\overrightarrow{AB}=\begin{pmatrix}x_B-x_A\y_B-y_A\end{pmatrix}

Two directed segments represent the same vector when they have the same horizontal and vertical changes, even if they start in different places. Reversing an arrow negates the vector: $\overrightarrow{BA}=-\overrightarrow{AB}$ . Segments can be chained head-to-tail, so $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ .

If $A=(-2,1)$ and $B=(3,4)$ , then $\overrightarrow{AB}=\begin{pmatrix}5\\3\end{pmatrix}$ . The segment from $C=(1,-4)$ to $D=(6,-1)$ has the same change $\begin{pmatrix}5\\3\end{pmatrix}$ , so it represents the same vector.

$\overrightarrow{AB}$ and $\overrightarrow{BA}$ lie on the same line but are not the same vector. A plain line segment has length but no chosen direction; the arrow and endpoint order are essential.

The position vector of a point $P$ is the vector from the origin to that point, written $\overrightarrow{OP}$ . It fixes the point's location, unlike a free vector that can be drawn with the same direction and length elsewhere.

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}

If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$ , then $\overrightarrow{AB}=\mathbf b-\mathbf a$ . To find an unknown position vector, begin at $O$ , follow any valid route to the point, and add the directed vectors on that route.

Let $M$ be the midpoint of $AB$ . Since $\overrightarrow{AM}=\tfrac12(\mathbf b-\mathbf a)$ , $\overrightarrow{OM}=\mathbf a+\tfrac12(\mathbf b-\mathbf a)=\tfrac12(\mathbf a+\mathbf b)$ . The coefficients add to 1 because $M$ lies on the line through $A$ and $B$ .

A point and its position vector are related but not identical: $P$ is the point, while $\overrightarrow{OP}$ is the directed displacement from the chosen origin. Do not write $\overrightarrow{AB}=\mathbf a-\mathbf b$ when $\mathbf a=\overrightarrow{OA}$ and $\mathbf b=\overrightarrow{OB}$ .

To express a vector in terms of two coplanar vectors, choose a complete route between its endpoints, replace each segment by a known vector or a fraction of one, then collect the two vector terms.

Use arrow direction before algebra: reverse a known segment by changing its sign; use equal opposite sides in a parallelogram; multiply a whole side by the required midpoint or division fraction; finally simplify coefficients of the same vector.

AP:PB=m:n\quad\Longrightarrow\quad\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}

Suppose $\overrightarrow{OA}=\mathbf a$ , $\overrightarrow{OB}=\mathbf b$ , and $M$ divides $AB$ in the ratio $AM:MB=2:3$ . Then $\overrightarrow{AB}=\mathbf b-\mathbf a$ and $\overrightarrow{AM}=\tfrac25(\mathbf b-\mathbf a)$ . Hence $\overrightarrow{OM}=\mathbf a+\overrightarrow{AM}=\tfrac35\mathbf a+\tfrac25\mathbf b$ .

The fraction attached to a segment is its share of the whole route: for $2:3$ , $AM$ is $2/5$ of $AB$ , not $2/3$ . Check the result by using a second route; both expressions must simplify to the same vector.

Vector geometry turns a diagram claim into an algebraic comparison. First derive two relevant vectors from clear routes; then factor them to expose a scalar multiple and state exactly what that proves.

Claim	Sufficient vector evidence	Required conclusion
Lines are parallel	$\mathbf u=k\mathbf v$ , $k\ne0$	same or opposite direction
$A,B,C$ are collinear	$\overrightarrow{AB}=k\overrightarrow{AC}$	both vectors share point $A$ and are parallel
$M$ is midpoint of $AB$	$\overrightarrow{AM}=\overrightarrow{MB}$	equal directed halves on the same line
$P$ divides $AB$ in $m:n$	$\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}$	state $AP:PB=m:n$

If $\overrightarrow{XY}=3\mathbf a+2\mathbf b$ and $\overrightarrow{ZY}=6\mathbf a+4\mathbf b$ , then $\overrightarrow{ZY}=2\overrightarrow{XY}$ . The vectors share endpoint $Y$ and are parallel in the same direction, so $Z,X,Y$ are collinear. Also $XY=\tfrac12 ZY$ , so $X$ is the midpoint of $ZY$ .

For ratio or similarity problems, express points as weighted position vectors and compare corresponding directed sides. A common non-zero scalar factor proves parallel directions and proportional lengths; use the diagram's shared points or side order to turn that fact into the requested ratio or similarity statement.

Equal magnitudes alone do not prove parallelism or collinearity. A proof must show a scalar-multiple relationship, preserve endpoint order, and finish with the geometric conclusion rather than stopping at the algebra.

Back to course directory

—Cambridge IGCSE Mathematics (0580)Read

E7.4 Vector geometry

Learning objectives

Represent a vector as a directed line segment

Use position vectors from the origin

Express a route using two coplanar vectors

Use vector multiples to prove geometric relationships

Back to course directory