E7.4 - Position vectors and geometric paths

E7.4 - Position vectors and geometric paths

Vector paths turn a diagram into algebra without needing to measure its angles. You will express directed journeys using two vectors, distinguish position vectors from displacements, and find points dividing a segment. These E7.4 methods prepare you to justify geometric relationships.

Directed paths and reverse vectors

Can two different routes give exactly the same displacement? If they start and finish at the same points, they must. The length travelled may change; the vector from start to finish does not.

Directed path

A directed line segment represents a vector by an arrow from its start to its finish. A path adds successive displacements head to tail. Reversing a segment negates its vector:

BA→=−AB→,AB→+BC→=AC→.\overrightarrow{BA}=-\overrightarrow{AB},\qquad \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}.

The middle point in ABAB then BCBC joins the two steps. You may choose any useful route between the same endpoints. A closed route has zero total displacement because it ends where it began. Here the vectors lie in one plane: they are coplanar.

[DIAGRAM: asset_name: Two vector paths in a parallelogram; asset_slug: parallelogram_paths; caption: A to O to B and A to C to B have the same start and finish.; description: Parallelogram O A C B has O at zero zero A at four zero B at one three C at five three. Directed arrows O to A and O to B are a and b. Directed arrows A to C and B to C are b and a. A dashed arrow A to B shows b minus a.]
A to O to B and A to C to B have the same start and finish.

Worked example 1: a path through the origin

Question. OACBOACB is a parallelogram, with OA→=a\overrightarrow{OA}=\mathbf{a} and OB→=b\overrightarrow{OB}=\mathbf{b}. Express AB→\overrightarrow{AB} using a\mathbf{a} and b\mathbf{b}, following the path through OO.

Plan. Travel from AA back to OO, then from OO to BB.

AB→=AO→+OB→join the two steps=−a+breverse OA=b−acollect the vectors\begin{aligned} \overrightarrow{AB}&=\overrightarrow{AO}+\overrightarrow{OB} && \text{join the two steps}\\ &=-\mathbf{a}+\mathbf{b} && \text{reverse }OA\\ &=\mathbf{b}-\mathbf{a} && \text{collect the vectors} \end{aligned}

Answer. AB→=b−a\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.

Check. Take a=(40)\mathbf{a}=\begin{pmatrix}4\\0\end{pmatrix}, b=(13)\mathbf{b}=\begin{pmatrix}1\\3\end{pmatrix}, as in the drawing. From A(4,0)A(4,0) to B(1,3)B(1,3), the displacement is (−33)\begin{pmatrix}-3\\3\end{pmatrix}, matching b−a\mathbf{b}-\mathbf{a}.

Keep the endpoints and diagram; change only the route to pass through CC.

Worked example 2: the other route

Question. Express AB→\overrightarrow{AB} using the path AA to CC to BB.

Plan. Use equal directed opposite sides of the parallelogram, reversing a side when necessary.

AC→=bsame displacement as OBCB→=−areverse the vector BCAB→=b−aadd along the path\begin{aligned} \overrightarrow{AC}&=\mathbf{b} && \text{same displacement as }OB\\ \overrightarrow{CB}&=-\mathbf{a} && \text{reverse the vector }BC\\ \overrightarrow{AB}&=\mathbf{b}-\mathbf{a} && \text{add along the path} \end{aligned}

Answer. AB→=b−a\overrightarrow{AB}=\mathbf{b}-\mathbf{a}, the same result as before.

Check. The closed path A→C→B→O→AA\to C\to B\to O\to A has total b−a−b+a=0\mathbf{b}-\mathbf{a}-\mathbf{b}+\mathbf{a}=\mathbf{0}, as a closed journey must.

Watch out

Equal parallel sides give equal vectors only when the directions match. In this parallelogram, AC→=OB→\overrightarrow{AC}=\overrightarrow{OB}, while CA→=−OB→\overrightarrow{CA}=-\overrightarrow{OB}.

Position vectors

A displacement says how to get from one point to another. A position vector fixes a point relative to a chosen origin, so every position arrow starts at that same origin.

Position vector

The position vector of AA, relative to origin OO, is OA→=a\overrightarrow{OA}=\mathbf{a}. If AA and BB have position vectors a\mathbf{a} and b\mathbf{b}, then

AB→=b−a.\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.

This is destination position minus start position, because the path A→O→BA\to O\to B is −a+b-\mathbf{a}+\mathbf{b}.

In Cartesian coordinates, the position vector of A(2,1)A(2,1) is (21)\begin{pmatrix}2\\1\end{pmatrix} when O=(0,0)O=(0,0). A column describing a displacement becomes a position vector only when its starting point is the chosen origin. If you change the origin, position vectors change, but the displacement between fixed points stays the same.

[DIAGRAM: asset_name: Position vectors and their difference; asset_slug: position_vector_difference; caption: The displacement from A to B is the difference of their position vectors.; description: Equal-scale grid shows O at zero zero A at two one and B at eight five. Blue arrow O to A is a, sage arrow O to B is b and clay arrow A to B is b minus a.]
The displacement from A to B is the difference of their position vectors.

Worked example 3: finding the destination position

Question. OA→=(21)\overrightarrow{OA}=\begin{pmatrix}2\\1\end{pmatrix} and AB→=(64)\overrightarrow{AB}=\begin{pmatrix}6\\4\end{pmatrix}. Find the position vector of BB.

Plan. Follow O→A→BO\to A\to B and add the two displacements.

OB→=OA→+AB→path from the origin=(21)+(64)add components=(85)destination position\begin{aligned} \overrightarrow{OB}&=\overrightarrow{OA}+\overrightarrow{AB} && \text{path from the origin}\\ &=\begin{pmatrix}2\\1\end{pmatrix}+\begin{pmatrix}6\\4\end{pmatrix} && \text{add components}\\ &=\begin{pmatrix}8\\5\end{pmatrix} && \text{destination position} \end{aligned}

Answer. OB→=(85)\overrightarrow{OB}=\begin{pmatrix}8\\5\end{pmatrix}.

Check. The coordinate displacement from (2,1)(2,1) to (8,5)(8,5) is (6,4)(6,4), the supplied ABAB vector.

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