Position vectors
- The position vector of a point A is \(\mathbf{a}=\overrightarrow{OA}\), where O is the origin.
- Its components are the coordinates of A. The point \(A(3,-1,2)\) has \(\mathbf{a}=\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}\).
AHL 4.5
Position vectors, displacement vectors and distance
AHL 4.5
Position vectors, displacement vectors and distance
By the end of this section
- I can write the position vector of a point from its coordinates.
- I can find the displacement vector AB between two points as b − a.
- I can say how AB and BA are related.
- I can find the exact distance between two points from their coordinates or position vectors.
Key points
Displacement vectors
- \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\): end minus start.
- To go from A to B, go back along \(\mathbf{a}\) to O, then out along \(\mathbf{b}\).
- \(\overrightarrow{BA}=-\overrightarrow{AB}\). The two have the same length and opposite directions.
- Write AB with the start first. \(\mathbf{a}-\mathbf{b}\) is \(\overrightarrow{BA}\), not \(\overrightarrow{AB}\).
Distance between two points
- The distance is the magnitude of the displacement: \(AB=|\overrightarrow{AB}|=|\mathbf{b}-\mathbf{a}|\).
- Subtract first, then square each component, add and take the square root.
- Leave an exact distance as a surd. In a context, give 3 significant figures with the unit.
Worked examples
Problems like the ones in class. Pick one, try it, then open the working.A displacement vector from coordinates
Points \(A(4, -2, 5)\) and \(B(1, 3, -1)\) are given. Find the displacement vector \(\overrightarrow{AB}\).
Working
Have a go first. Then open the working one line at a time.
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Position vectors
\(\mathbf{a}=\begin{pmatrix}4 \\ -2 \\ 5\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}1 \\ 3 \\ -1\end{pmatrix}\)
The components of a position vector are the coordinates of the point.
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End minus start
\(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}1 \\ 3 \\ -1\end{pmatrix}-\begin{pmatrix}4 \\ -2 \\ 5\end{pmatrix}\)
From A go back to O, then out to B.
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Subtract each component
\(\overrightarrow{AB}=\begin{pmatrix}-3 \\ 5 \\ -6\end{pmatrix}\)
Take each component of B minus the matching one of A.
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The other way
\(\overrightarrow{BA}=-\overrightarrow{AB}=\begin{pmatrix}3 \\ -5 \\ 6\end{pmatrix}\)
BA has the same length as AB and the opposite direction.
An exact distance
Points A and B have position vectors \(\mathbf{a}=5\mathbf{i}-2\mathbf{j}+\mathbf{k}\) and \(\mathbf{b}=2\mathbf{i}+4\mathbf{j}-3\mathbf{k}\). Find the exact distance \(AB\).
Working
Have a go first. Then open the working one line at a time.
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Displacement
\(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(2\mathbf{i}+4\mathbf{j}-3\mathbf{k})-(5\mathbf{i}-2\mathbf{j}+\mathbf{k})=-3\mathbf{i}+6\mathbf{j}-4\mathbf{k}\)
Subtract coefficient by coefficient.
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Magnitude
\(AB=|\overrightarrow{AB}|=\sqrt{(-3)^2+6^2+(-4)^2}\)
The distance is the magnitude of the displacement.
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Exact answer
\(AB=\sqrt{61}\)
Leave it as a surd. 61 has no square factor, so it does not simplify.
Distance between two drones
Relative to a base at O, drone P is at \(\begin{pmatrix}30 \\ -10 \\ 12\end{pmatrix}\) and drone Q is at \(\begin{pmatrix}-6 \\ 14 \\ 40\end{pmatrix}\), in metres, where \(\mathbf{k}\) is vertically up. Find the vector \(\overrightarrow{QP}\) and the distance between the drones to 3 significant figures.
Working
Have a go first. Then open the working one line at a time.
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QP
\(\overrightarrow{QP}=\mathbf{p}-\mathbf{q}=\begin{pmatrix}30 \\ -10 \\ 12\end{pmatrix}-\begin{pmatrix}-6 \\ 14 \\ 40\end{pmatrix}=\begin{pmatrix}36 \\ -24 \\ -28\end{pmatrix}\)
QP is from Q to P: end P minus start Q.
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Distance
\(PQ=\sqrt{(-36)^2+24^2+28^2}=\sqrt{2656}=51.5364\)
PQ and QP are the same length.
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Round
\(PQ=51.5\text{ m}\)
In a context, give 3 significant figures and the unit.