10.5 Vectors in pure mathematics and in context — Practise with help

Work each one first, closed book. For a proof, write down the two vectors you will compare. For a motion or force problem, sketch the situation and choose axes before you add anything.

Warm-up: remember

Solve $3x - 4 = 11$ .
Find the midpoint of $A(2, 1)$ and $B(8, 5)$ .
Find the magnitude of $4\mathbf{i} - 3\mathbf{j}$ .

Answers

Add $4$ to both sides: $3x = 15$ , so $x = 5$ .
The midpoint is $\left(\dfrac{2 + 8}{2}, \dfrac{1 + 5}{2}\right) = (5, 3)$ .
$\sqrt{4^2 + (-3)^2} = \sqrt{25} = 5$ .

If question 1 was hard, revise solving linear equations before you continue.

Practise with help

1. Show that $A(0, 1)$ , $B(2, 5)$ and $C(4, 9)$ are collinear.

Hint

Find $\vec{AB}$ and $\vec{BC}$ . If one is a multiple of the other, and they share $B$ , the points are collinear.

Next step

$\vec{AB} = (2 - 0)\mathbf{i} + (5 - 1)\mathbf{j}$ . Work out $\vec{BC}$ the same way and compare.

Full working

\vec{AB} = 2\mathbf{i} + 4\mathbf{j}, \qquad \vec{BC} = 2\mathbf{i} + 4\mathbf{j}.

So $\vec{AB} = \vec{BC}$ , and the two vectors share the point $B$ . The three points lie on one straight line.

2. The quadrilateral has vertices $A(1, 1)$ , $B(4, 1)$ , $C(4, 4)$ and $D(1, 4)$ . Show that it is a parallelogram.

Hint

A quadrilateral is a parallelogram when one pair of opposite sides is equal and parallel. Compare $\vec{AB}$ with $\vec{DC}$ .

Next step

$\vec{AB} = (4 - 1)\mathbf{i} + (1 - 1)\mathbf{j}$ . For $\vec{DC}$ , subtract $D$ from $C$ .

Full working

\vec{AB} = 3\mathbf{i}, \qquad \vec{DC} = (4 - 1)\mathbf{i} + (4 - 4)\mathbf{j} = 3\mathbf{i}.

So $\vec{AB} = \vec{DC}$ . One pair of opposite sides is equal and parallel, so $ABCD$ is a parallelogram.

3. The points are $A(0, 0)$ and $B(6, 3)$ . Find the point $P$ on $AB$ with $AP : PB = 2 : 1$ .

Hint

Use $\vec{OP} = \dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}$ with $m = 2$ and $n = 1$ .

Next step

Work out $\dfrac{1 \times \mathbf{a} + 2 \times \mathbf{b}}{3}$ .

Full working

\vec{OP} = \frac{\mathbf{a} + 2\mathbf{b}}{3} = \frac{0 + 2(6\mathbf{i} + 3\mathbf{j})}{3} = \frac{12\mathbf{i} + 6\mathbf{j}}{3} = 4\mathbf{i} + 2\mathbf{j}.

So $P$ is $(4, 2)$ . Check: $\vec{AP} = 4\mathbf{i} + 2\mathbf{j}$ and $\vec{PB} = 2\mathbf{i} + \mathbf{j}$ , so $AP : PB = 2 : 1$ .

4. Forces $\mathbf{F}_1 = 2\mathbf{i} + 5\mathbf{j}$ N and $\mathbf{F}_2 = -3\mathbf{i} + \mathbf{j}$ N act at a point. Find the force $\mathbf{F}_3$ that keeps the point still, and its magnitude.

Hint

In equilibrium the three forces sum to zero, so $\mathbf{F}_3$ is the negative of the sum of the first two.

Next step

$\mathbf{F}_1 + \mathbf{F}_2 = -\mathbf{i} + 6\mathbf{j}$ . Change each sign.

Full working

\mathbf{F}_1 + \mathbf{F}_2 = (2 - 3)\mathbf{i} + (5 + 1)\mathbf{j} = -\mathbf{i} + 6\mathbf{j},

\mathbf{F}_3 = -(-\mathbf{i} + 6\mathbf{j}) = \mathbf{i} - 6\mathbf{j} \text{ N}.

|\mathbf{F}_3| = \sqrt{1^2 + (-6)^2} = \sqrt{37} \approx 6.08 \text{ N}.

5. A particle starts at position vector $\mathbf{r}_0 = \mathbf{i} + 2\mathbf{j}$ m and moves with velocity $\mathbf{v} = 2\mathbf{i} + 3\mathbf{j}$ m/s. Find its position after $5$ s and its distance from the origin then.

Hint

Use $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t$ . Multiply the velocity by the time before you add.

Next step

$5\mathbf{v} = 10\mathbf{i} + 15\mathbf{j}$ . Add this to $\mathbf{r}_0$ .

Full working

\mathbf{r} = (\mathbf{i} + 2\mathbf{j}) + 5(2\mathbf{i} + 3\mathbf{j}) = 11\mathbf{i} + 17\mathbf{j}.

The position is $(11, 17)$ m. The distance from the origin is

|\mathbf{r}| = \sqrt{11^2 + 17^2} = \sqrt{121 + 289} = \sqrt{410} \approx 20.2 \text{ m}.

6. A boat moves at $4\mathbf{i}$ m/s through still water. A current flows at $-3\mathbf{j}$ m/s. Find the resultant velocity, the speed and the direction from the positive $x$ -axis.

Hint

Add the two velocities first. The speed is the magnitude of the resultant, and the direction comes from its components.

Next step

The resultant is $4\mathbf{i} - 3\mathbf{j}$ . Find its length, then use the signs to place the direction below the positive $x$ -axis.

Full working

\mathbf{v} = 4\mathbf{i} - 3\mathbf{j}, \qquad |\mathbf{v}| = \sqrt{4^2 + (-3)^2} = 5 \text{ m/s}.

The direction is $\tan^{-1}\left(\frac{3}{4}\right) = 36.9^\circ$ below the positive $x$ -axis, which is $-36.9^\circ$ or $323.1^\circ$ anticlockwise.

7. A plane flies at $150$ km/h due east. A wind of $40$ km/h blows from the north, so it pushes the plane due south. Find the resultant velocity, the speed and the direction.

Hint

Take east as $\mathbf{i}$ . The plane gives $150\mathbf{i}$ , and the wind gives $-40\mathbf{j}$ .

Next step

Work out $\sqrt{150^2 + 40^2}$ and $\tan^{-1}\left(\frac{40}{150}\right)$ .

Full working

\mathbf{v} = 150\mathbf{i} - 40\mathbf{j}, \qquad |\mathbf{v}| = \sqrt{150^2 + 40^2} = \sqrt{24100} \approx 155.2 \text{ km/h}.

The direction is $\tan^{-1}\left(\frac{40}{150}\right) = 14.9^\circ$ south of east. The wind has pushed the plane off its due-east course.

8. Show that $A(1, 1)$ , $B(3, 5)$ and $C(6, 11)$ are collinear.

Hint

This time the multiplier is not $1$ . Show that $\vec{BC}$ is a multiple of $\vec{AB}$ , and name the shared point.

Next step

$\vec{AB} = 2\mathbf{i} + 4\mathbf{j}$ and $\vec{BC} = 3\mathbf{i} + 6\mathbf{j}$ . What number multiplies $\vec{AB}$ to give $\vec{BC}$ ?

Full working

\vec{AB} = 2\mathbf{i} + 4\mathbf{j}, \qquad \vec{BC} = 3\mathbf{i} + 6\mathbf{j} = \frac{3}{2}\left(2\mathbf{i} + 4\mathbf{j}\right) = \frac{3}{2}\vec{AB}.

So $\vec{BC}$ is a positive multiple of $\vec{AB}$ . The two vectors share the point $B$ , so $A$ , $B$ and $C$ are collinear.