7D, 7F - Position vectors and division points

7D, 7F - Position vectors and division points

Two places can be described by arrows from one origin, but how do we find the movement from one place to the other? You will use position vectors to join points, locate a point dividing a segment in a given ratio and calculate the same vector by different routes through a figure. In 7D, you describe positions relative to an origin; in 7F, you locate points dividing a segment.

Vectors from an origin

Choosing one fixed origin lets us describe each point by the displacement needed to reach it.

Position vector

The position vector of AA, relative to origin OO, is OA→=a\overrightarrow{OA}=\mathbf{a}. It fixes the location of AA once the origin is chosen. The point OO has position vector 0\mathbf{0}.

If OB→=b\overrightarrow{OB}=\mathbf{b}, the route OO to AA to BB gives a+AB→=b\mathbf{a}+\overrightarrow{AB}=\mathbf{b}. Subtracting a\mathbf{a} isolates the movement from AA to BB.

Joining two position vectors

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

Use finishing position minus starting position. Equivalently, travel from AA back to OO, then from OO to BB.

The origin is a reference point, not a compulsory stop on the actual journey. If we move the origin, the two position vectors change, but their difference still represents the same displacement AB→\overrightarrow{AB}.

Worked example 1: finding the joining vector

Question. Given OA→=2i+3j\overrightarrow{OA}=2\mathbf{i}+3\mathbf{j} and OB→=7i−j\overrightarrow{OB}=7\mathbf{i}-\mathbf{j}, find AB→\overrightarrow{AB}.

Plan. Subtract the position of the starting point from the position of the finishing point.

AB→=(7i−j)−(2i+3j)=5i−4j\begin{aligned} \overrightarrow{AB}&=(7\mathbf{i}-\mathbf{j})-(2\mathbf{i}+3\mathbf{j})\\ &=5\mathbf{i}-4\mathbf{j} \end{aligned}

Answer. AB→=5i−4j\overrightarrow{AB}=5\mathbf{i}-4\mathbf{j}.

Check. Starting at (2,3)(2,3), moving 55 right and 44 down reaches (7,−1)(7,-1), the position of BB.

Now the movement is known and the finishing position is missing. Rearrange the same route relation.

Worked example 2: finding a new position

Question. Given OA→=2i+3j\overrightarrow{OA}=2\mathbf{i}+3\mathbf{j} and AB→=−4i+5j\overrightarrow{AB}=-4\mathbf{i}+5\mathbf{j}, find OB→\overrightarrow{OB}.

Plan. Add the displacement from AA to the position vector of AA.

OB→=OA→+AB→=(2i+3j)+(−4i+5j)=−2i+8j\begin{aligned} \overrightarrow{OB}&=\overrightarrow{OA}+\overrightarrow{AB}\\ &=(2\mathbf{i}+3\mathbf{j})+(-4\mathbf{i}+5\mathbf{j})\\ &=-2\mathbf{i}+8\mathbf{j} \end{aligned}

Answer. OB→=−2i+8j\overrightarrow{OB}=-2\mathbf{i}+8\mathbf{j}.

Check. Subtracting OA→\overrightarrow{OA} from the answer gives −4i+5j-4\mathbf{i}+5\mathbf{j}, the supplied movement.

Read the letter order before subtracting.

A point dividing a segment

A point partway along ABAB can be reached by starting at AA and taking a fraction of the displacement AB→\overrightarrow{AB}.

Suppose PP divides the segment internally with AP:PB=m:nAP:PB=m:n, where m,n>0m,n>0. There are m+nm+n equal ratio parts in the whole segment, and APAP contains mm of them. Therefore AP→=mm+nAB→\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}. The fraction lies strictly between 00 and 11, which places PP between the endpoints.

Internal division

For distinct A,BA,B, with OA→=a\overrightarrow{OA}=\mathbf{a}, OB→=b\overrightarrow{OB}=\mathbf{b} and AP:PB=m:nAP:PB=m:n, where m,n>0m,n>0,

OP→=a+mm+n(b−a)=na+mbm+n\begin{aligned} \overrightarrow{OP}&=\mathbf{a}+\frac{m}{m+n}(\mathbf{b}-\mathbf{a})\\ &=\frac{n\mathbf{a}+m\mathbf{b}}{m+n} \end{aligned}

For a midpoint, m=nm=n, giving OP→=(a+b)/2\overrightarrow{OP}=(\mathbf{a}+\mathbf{b})/2.

Notice that the weight of b\mathbf{b} is the proportion travelled from AA towards BB. Deriving the route prevents the common mistake of putting the ratio weights beside the wrong endpoints.

[DIAGRAM: asset_name: Internal division by a vector route; asset_slug: internal-vector-division; caption: A point two thirds of the way from A to B has position vector one third a plus two thirds b; description: Triangle OAB has position vectors a equals 6i plus j and b equals minus 2i plus 5j. P lies on AB with AP to PB equal to 2 to 1. The arrow OP locates the point with position vector one third a plus two thirds b.]
A point two thirds of the way from A to B has position vector one third a plus two thirds b

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