10.4 Position vectors and the distance between two points — Practise with help

Work each one first, closed book. Before you calculate, decide whether the question wants a displacement vector, a distance or a midpoint. Write the subtraction in the order the letters give.

Warm-up: remember

Write down $5 - 1$ and $1 - 5$ .
Find the magnitude of $3\mathbf{i} + 4\mathbf{j}$ .
Work out the midpoint of $2$ and $8$ .

Answers

$5 - 1 = 4$ and $1 - 5 = -4$ .
$\sqrt{3^2 + 4^2} = 5$ .
The midpoint is $\dfrac{2 + 8}{2} = 5$ .

If question 1 was hard, revise subtracting directed numbers before you continue.

Practise with help

1. Write down the position vector of $A(3, -2)$ .

Hint

The position vector of a point runs from the origin to the point. Its components are the coordinates.

Next step

Read the coordinates into a column: $3$ on top, $-2$ below.

Full working

\vec{OA} = 3\mathbf{i} - 2\mathbf{j} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}.

2. The points are $A(2, 1)$ and $B(7, 4)$ . Find $\vec{AB}$ .

Hint

$\vec{AB}$ is the position vector of $B$ minus the position vector of $A$ . The letter that comes second is the one you subtract from.

Next step

Work out $7 - 2$ and $4 - 1$ .

Full working

\vec{AB} = \mathbf{b} - \mathbf{a} = (7 - 2)\mathbf{i} + (4 - 1)\mathbf{j} = 5\mathbf{i} + 3\mathbf{j}.

3. Find the distance between $A(1, -1)$ and $B(4, 3)$ .

Hint

First find $\vec{AB}$ , then find its magnitude.

Next step

$\vec{AB} = (4 - 1)\mathbf{i} + (3 - (-1))\mathbf{j} = 3\mathbf{i} + 4\mathbf{j}$ . Now take the square root of the sum of squares.

Full working

\vec{AB} = 3\mathbf{i} + 4\mathbf{j}, \qquad AB = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.

4. Find the distance between $A(2, 0, 1)$ and $B(5, 4, 3)$ .

Hint

There are three coordinates, so there are three squares under the square root.

Next step

The difference of the $z$ -coordinates is $3 - 1 = 2$ .

Full working

\vec{AB} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}, \qquad AB = \sqrt{3^2 + 4^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29}.

The distance is $\sqrt{29} \approx 5.39$ .

5. Find the distance of $P(-3, 4, 12)$ from the origin.

Hint

The distance from the origin is the magnitude of the position vector.

Next step

Square each coordinate, including $(-3)^2$ and $12^2$ , then add.

Full working

OP = \sqrt{(-3)^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13.

6. Find the midpoint of $A(1, 3)$ and $B(7, -1)$ .

Hint

The midpoint is the average of the two position vectors. Average each coordinate separately.

Next step

Work out $\dfrac{1 + 7}{2}$ and $\dfrac{3 + (-1)}{2}$ .

Full working

\mathbf{m} = \frac{1}{2}\left((1 + 7)\mathbf{i} + (3 - 1)\mathbf{j}\right) = 4\mathbf{i} + \mathbf{j}.

The midpoint is $(4, 1)$ .

7. Show that the triangle with vertices $A(1, 1)$ , $B(4, 5)$ and $C(8, 2)$ is isosceles.

Hint

An isosceles triangle has two equal sides. Work out all three side lengths and look for a matching pair.

Next step

$AB$ uses $\vec{AB} = 3\mathbf{i} + 4\mathbf{j}$ and $BC$ uses $\vec{BC} = 4\mathbf{i} - 3\mathbf{j}$ .

Full working

AB = \sqrt{3^2 + 4^2} = 5, \qquad BC = \sqrt{4^2 + (-3)^2} = 5,

AC = \sqrt{7^2 + 1^2} = \sqrt{50}.

Two sides are equal, so the triangle is isosceles.

8. A point $P$ has position vector $2\mathbf{i} - \mathbf{j}$ and a point $Q$ has position vector $6\mathbf{i} + 2\mathbf{j}$ . Find $\vec{PQ}$ and the distance $PQ$ .

Hint

Subtract the position vector of $P$ from the position vector of $Q$ , then take the magnitude.

Next step

$\vec{PQ} = (6 - 2)\mathbf{i} + (2 - (-1))\mathbf{j}$ .

Full working

\vec{PQ} = 4\mathbf{i} + 3\mathbf{j}, \qquad PQ = \sqrt{4^2 + 3^2} = 5.