10.4 Position vectors and the distance between two points — Practise

Work through these on your own. State the origin for a position vector, keep the subtraction in the order the letters give, and finish with a magnitude when the question asks for a distance.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Write down the position vector of $A(-2, 5)$ from the origin.
For $A(1, 3)$ and $B(6, 11)$ , find $\vec{AB}$ .
Find the distance between $O(0, 0)$ and $B(9, 12)$ .
Find the distance between $A(-2, 1)$ and $B(4, 9)$ .
Find the distance between $A(1, -2, 3)$ and $B(4, 2, 5)$ .
Find the distance of $P(2, -3, 6)$ from the origin.
Find the midpoint of $A(-1, 4)$ and $B(5, -2)$ .
Find the midpoint of $A(2, 1, -4)$ and $B(6, -3, 2)$ .
The points are $A(1, 2)$ , $B(4, 6)$ and $C(7, 10)$ . Find $AB$ , $BC$ and $AC$ , and explain what the three distances show.
The distance between $A(2, 1)$ and $B(k, 5)$ is $5$ . Find the two possible values of $k$ .
The points are $A(0, 0)$ , $B(3, 4)$ and $C(7, 1)$ . Show that the angle at $B$ is a right angle.
Find the distance between $A(1, 1, 1)$ and $B(3, 5, 13)$ .

Answers and concise solutions

1. The position vector

Answer: The position vector of a point has the coordinates as its components: $\vec{OA} = -2\mathbf{i} + 5\mathbf{j}$ , the column $\begin{pmatrix} -2 \\ 5 \end{pmatrix}$ .

2. The displacement

Answer: $\vec{AB} = \mathbf{b} - \mathbf{a} = (6 - 1)\mathbf{i} + (11 - 3)\mathbf{j} = 5\mathbf{i} + 8\mathbf{j}$ . The letter that comes second is the one you subtract from.

3. The distance

Answer: $OB = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$ .

4. The distance

Answer: $\vec{AB} = (4 - (-2))\mathbf{i} + (9 - 1)\mathbf{j} = 6\mathbf{i} + 8\mathbf{j}$ , so $AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ .

5. The distance

Answer: $\vec{AB} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}$ , so $AB = \sqrt{3^2 + 4^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29} \approx 5.39$ .

6. The distance

Answer: The distance from the origin is the magnitude of the position vector: $\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ .

7. The midpoint

Answer: The midpoint is the average of the position vectors: $\dfrac{1}{2}\big((-1 + 5)\mathbf{i} + (4 - 2)\mathbf{j}\big) = \dfrac{1}{2}(4\mathbf{i} + 2\mathbf{j}) = 2\mathbf{i} + \mathbf{j}$ , which is the point $(2, 1)$ .

8. The midpoint

Answer: Average each coordinate: $\dfrac{1}{2}(8\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}) = 4\mathbf{i} - \mathbf{j} - \mathbf{k}$ . The midpoint is $(4, -1, -1)$ .

9. The three distances

Answer: $\vec{AB} = 3\mathbf{i} + 4\mathbf{j}$ gives $AB = 5$ , and $\vec{BC} = 3\mathbf{i} + 4\mathbf{j}$ gives $BC = 5$ . Then $\vec{AC} = 6\mathbf{i} + 8\mathbf{j}$ gives $AC = 10$ . Since $AB + BC = 5 + 5 = 10 = AC$ , the two shorter distances add to the longest, so the three points lie on a straight line.

10. The values of $k$

Answer: $\vec{AB} = (k - 2)\mathbf{i} + 4\mathbf{j}$ , so $AB^2 = (k - 2)^2 + 16 = 25$ . Then $(k - 2)^2 = 9$ , so $k - 2 = 3$ or $k - 2 = -3$ . The two values are $k = 5$ and $k = -1$ .

11. The check

Answer: $AB = \sqrt{3^2 + 4^2} = 5$ and $BC = \sqrt{4^2 + (-3)^2} = 5$ , while $AC = \sqrt{7^2 + 1^2} = \sqrt{50}$ . Since $AB^2 + BC^2 = 25 + 25 = 50 = AC^2$ , the triangle satisfies Pythagoras' theorem, so the angle at $B$ is a right angle.

12. The distance

Answer: $\vec{AB} = 2\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$ , so $AB = \sqrt{2^2 + 4^2 + 12^2} = \sqrt{4 + 16 + 144} = \sqrt{164} = 2\sqrt{41} \approx 12.81$ .