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

Points \(A(3, -5, 2)\) and \(B(-1, 4, 7)\) are given. Find:

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

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}-1 \\ 4 \\ 7\end{pmatrix}-\begin{pmatrix}3 \\ -5 \\ 2\end{pmatrix}=\begin{pmatrix}-4 \\ 9 \\ 5\end{pmatrix}\)

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

   • \(\overrightarrow{BA}=-\overrightarrow{AB}=\begin{pmatrix}4 \\ -9 \\ -5\end{pmatrix}\)

Points A and B have position vectors \(\mathbf{a}=2\mathbf{i}+7\mathbf{j}-3\mathbf{k}\) and \(\mathbf{b}=5\mathbf{i}-\mathbf{j}+4\mathbf{k}\). Write \(\overrightarrow{AB}\) in the form \(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\).

• \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(5\mathbf{i}-\mathbf{j}+4\mathbf{k})-(2\mathbf{i}+7\mathbf{j}-3\mathbf{k})=3\mathbf{i}-8\mathbf{j}+7\mathbf{k}\)

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

1. (a) \(\mathbf{a}=3\mathbf{i}-2\mathbf{j}+\mathbf{k},\ \mathbf{b}=\mathbf{i}+4\mathbf{j}-5\mathbf{k};\)

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(\mathbf{i}+4\mathbf{j}-5\mathbf{k})-(3\mathbf{i}-2\mathbf{j}+\mathbf{k})=-2\mathbf{i}+6\mathbf{j}-6\mathbf{k}\)
   • \(AB=\sqrt{(-2)^2+6^2+(-6)^2}=2 \sqrt{19}\)

2. (b) \(\mathbf{a}=2\mathbf{i}+\mathbf{j},\ \mathbf{b}=5\mathbf{i}-3\mathbf{j}+\mathbf{k}.\)

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(5\mathbf{i}-3\mathbf{j}+\mathbf{k})-(2\mathbf{i}+\mathbf{j})=3\mathbf{i}-4\mathbf{j}+\mathbf{k}\)
   • \(AB=\sqrt{3^2+(-4)^2+1^2}=\sqrt{26}\)

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

1. (a) \(\mathbf{a}=\begin{pmatrix}2 \\ -1 \\ 3\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}-2 \\ 1 \\ -1\end{pmatrix};\)

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

2. (b) \(\mathbf{a}=\begin{pmatrix}5 \\ 0 \\ -2\end{pmatrix},\ \mathbf{b}=\begin{pmatrix}1 \\ 3 \\ -1\end{pmatrix}.\)

   • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}1 \\ 3 \\ -1\end{pmatrix}-\begin{pmatrix}5 \\ 0 \\ -2\end{pmatrix}=\begin{pmatrix}-4 \\ 3 \\ 1\end{pmatrix}\)
   • \(AB=\sqrt{(-4)^2+3^2+1^2}=\sqrt{26}\)

Relative to a control tower at O, drone P has position vector \(\mathbf{p}=\begin{pmatrix}40 \\ -25 \\ 60\end{pmatrix}\) and drone Q has position vector \(\mathbf{q}=\begin{pmatrix}-10 \\ 30 \\ 20\end{pmatrix}\), in metres, where \(\mathbf{k}\) is vertically up. Find the distance \(PQ\) between the drones to 3 significant figures.

• \(\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}=\begin{pmatrix}-50 \\ 55 \\ -40\end{pmatrix}\)
• \(PQ=\sqrt{(-50)^2+55^2+(-40)^2}=\sqrt{7125}=84.410\ldots\)
• To 3 significant figures: 84.4 m.

A hiker starts at the point \(A(2,-3,5)\) and walks to a point B. The displacement is \(\overrightarrow{AB}=\begin{pmatrix}4 \\ 1 \\ -6\end{pmatrix}\). Find the position vector of B.

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

The points \(A(3, -2, 4)\), \(B(-3, -3, 2)\) and \(C(2, 4, 6)\) are the corners of a triangle.

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

   • \(\overrightarrow{AB}=\begin{pmatrix}-6 \\ -1 \\ -2\end{pmatrix},\ AB=\sqrt{41}\)
   • \(\overrightarrow{AC}=\begin{pmatrix}-1 \\ 6 \\ 2\end{pmatrix},\ AC=\sqrt{41}\)
   • \(\overrightarrow{BC}=\begin{pmatrix}5 \\ 7 \\ 4\end{pmatrix},\ BC=\sqrt{90}\)

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

   • \(AB=AC=\sqrt{41}\)
   • Two sides are equal, so the triangle is isosceles with apex A.

The points \(A(3, 6, 2)\), \(B(-3, 5, 0)\) and \(C(-1, -3, -2)\) are the corners of a triangle.

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

   • \(\overrightarrow{AB}=\begin{pmatrix}-6 \\ -1 \\ -2\end{pmatrix},\ AB^2=41\)
   • \(\overrightarrow{BC}=\begin{pmatrix}2 \\ -8 \\ -2\end{pmatrix},\ BC^2=72\)
   • \(\overrightarrow{AC}=\begin{pmatrix}-4 \\ -9 \\ -4\end{pmatrix},\ AC^2=113\)

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

   • \(AB^2+BC^2=41+72=113=AC^2\)
   • The sides satisfy Pythagoras, so the angle at B is \(90^\circ\).