Position vectors, displacement vectors and distance Problem set B Worked solutions

Points \(A(-4, 2, 6)\) and \(B(5, -3, 1)\) are given. Find:

1. (a) \(\overrightarrow{AB}\);

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}5 \\ -3 \\ 1\end{pmatrix}-\begin{pmatrix}-4 \\ 2 \\ 6\end{pmatrix}=\begin{pmatrix}9 \\ -5 \\ -5\end{pmatrix}\)

2. (b) \(\overrightarrow{BA}\).

   • \(\overrightarrow{BA}=-\overrightarrow{AB}=\begin{pmatrix}-9 \\ 5 \\ 5\end{pmatrix}\)

Points A and B have position vectors \(\mathbf{a}=-3\mathbf{i}+4\mathbf{j}+\mathbf{k}\) and \(\mathbf{b}=\mathbf{i}+\mathbf{j}-6\mathbf{k}\). Write \(\overrightarrow{AB}\) in the form \(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\).

• \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(\mathbf{i}+\mathbf{j}-6\mathbf{k})-(-3\mathbf{i}+4\mathbf{j}+\mathbf{k})=4\mathbf{i}-3\mathbf{j}-7\mathbf{k}\)

Find the exact distance \(AB\) between points A and B with position vectors:

1. (a) \(\mathbf{a}=4\mathbf{i}-\mathbf{j}+2\mathbf{k},\ \mathbf{b}=5\mathbf{j}-2\mathbf{k};\)

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(5\mathbf{j}-2\mathbf{k})-(4\mathbf{i}-\mathbf{j}+2\mathbf{k})=-4\mathbf{i}+6\mathbf{j}-4\mathbf{k}\)
   • \(AB=\sqrt{(-4)^2+6^2+(-4)^2}=2 \sqrt{17}\)

2. (b) \(\mathbf{a}=\mathbf{i}+3\mathbf{j},\ \mathbf{b}=3\mathbf{i}-\mathbf{j}+4\mathbf{k}.\)

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(3\mathbf{i}-\mathbf{j}+4\mathbf{k})-(\mathbf{i}+3\mathbf{j})=2\mathbf{i}-4\mathbf{j}+4\mathbf{k}\)
   • \(AB=\sqrt{2^2+(-4)^2+4^2}=6\)

Find the exact distance \(AB\) between points A and B with position vectors:

1. (a) \(\mathbf{a}=\begin{pmatrix}1 \\ 4 \\ -2\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}7 \\ 2 \\ 1\end{pmatrix};\)

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}7 \\ 2 \\ 1\end{pmatrix}-\begin{pmatrix}1 \\ 4 \\ -2\end{pmatrix}=\begin{pmatrix}6 \\ -2 \\ 3\end{pmatrix}\)
   • \(AB=\sqrt{6^2+(-2)^2+3^2}=7\)

2. (b) \(\mathbf{a}=\begin{pmatrix}0 \\ -3 \\ 5\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}2 \\ -6 \\ 6\end{pmatrix}.\)

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}2 \\ -6 \\ 6\end{pmatrix}-\begin{pmatrix}0 \\ -3 \\ 5\end{pmatrix}=\begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix}\)
   • \(AB=\sqrt{2^2+(-3)^2+1^2}=\sqrt{14}\)

Relative to the corner O of a fish tank, a goldfish P has position vector \(\mathbf{p}=\begin{pmatrix}18 \\ 22 \\ 7\end{pmatrix}\) and a snail Q has position vector \(\mathbf{q}=\begin{pmatrix}-6 \\ 10 \\ 35\end{pmatrix}\), in centimetres, where \(\mathbf{k}\) is vertically up. Find the distance \(PQ\) between them to 3 significant figures.

• \(\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=\begin{pmatrix}-24 \\ -12 \\ 28\end{pmatrix}\)
• \(PQ=\sqrt{(-24)^2+(-12)^2+28^2}=\sqrt{1504}=38.781\ldots\)
• To 3 significant figures: 38.8 cm.

A skier starts at the point \(A(-5,8,2)\) and ends at a point B. The displacement is \(\overrightarrow{AB}=\begin{pmatrix}3 \\ -4 \\ 9\end{pmatrix}\). Find the position vector of B.

• \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\ \text{so}\ \mathbf{b}=\mathbf{a}+\overrightarrow{AB}\)
• \(\mathbf{b}=\begin{pmatrix}-5 \\ 8 \\ 2\end{pmatrix}+\begin{pmatrix}3 \\ -4 \\ 9\end{pmatrix}=\begin{pmatrix}-2 \\ 4 \\ 11\end{pmatrix}\)

The points \(A(3, 1, 1)\), \(B(0, 4, 5)\) and \(C(-2, -2, 1)\) are the corners of a triangle.

1. (a) Find the exact lengths \(AB\), \(AC\) and \(BC\).

   • \(\overrightarrow{AB}=\begin{pmatrix}-3 \\ 3 \\ 4\end{pmatrix},\ AB=\sqrt{34}\)
   • \(\overrightarrow{AC}=\begin{pmatrix}-5 \\ -3 \\ 0\end{pmatrix},\ AC=\sqrt{34}\)
   • \(\overrightarrow{BC}=\begin{pmatrix}-2 \\ -6 \\ -4\end{pmatrix},\ BC=\sqrt{56}\)

2. (b) Show that the triangle is isosceles.

   • \(AB=AC=\sqrt{34}\)
   • Two sides are equal, so the triangle is isosceles with apex A.

The points \(A(2, 3, 2)\), \(B(-2, 4, 0)\) and \(C(0, 4, -4)\) are the corners of a triangle.

1. (a) Find \(AB^2\), \(BC^2\) and \(AC^2\).

   • \(\overrightarrow{AB}=\begin{pmatrix}-4 \\ 1 \\ -2\end{pmatrix},\ AB^2=21\)
   • \(\overrightarrow{BC}=\begin{pmatrix}2 \\ 0 \\ -4\end{pmatrix},\ BC^2=20\)
   • \(\overrightarrow{AC}=\begin{pmatrix}-2 \\ 1 \\ -6\end{pmatrix},\ AC^2=41\)

2. (b) Show that angle \(ABC\) is a right angle.

   • \(AB^2+BC^2=21+20=41=AC^2\)
   • The sides satisfy Pythagoras, so the angle at B is \(90^\circ\).