10.3 Vector addition and scalar multiplication — Practise with help

Work each one first, closed book. Decide whether the question wants a vector, a magnitude or a direction. Add or scale the components before you take any square root.

Warm-up: remember

Work out $3 \times 4$ and $3 \times (-2)$ .
Work out $(2 + 5, 3 - 1)$ .
Find the magnitude of $6\mathbf{i} + 8\mathbf{j}$ .

Answers

$3 \times 4 = 12$ and $3 \times (-2) = -6$ .
Add the matching positions: $(2 + 5, 3 - 1) = (7, 2)$ .
$\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ .

If question 2 was hard, revise adding directed numbers before you continue.

Practise with help

1. Add $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$ and $\mathbf{b} = 5\mathbf{i} - \mathbf{j}$ .

Hint

Add the coefficients of $\mathbf{i}$ together, and the coefficients of $\mathbf{j}$ together.

Next step

The $x$ -component is $2 + 5$ and the $y$ -component is $3 + (-1)$ .

Full working

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

The resultant is $7\mathbf{i} + 2\mathbf{j}$ .

2. Add $\mathbf{a} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ .

Hint

There are three rows in each column. Add each row on its own, and keep the signs with the numbers.

Next step

For the $\mathbf{k}$ row, work out $-1 + 4$ .

Full working

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

3. Multiply $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j}$ by $4$ .

Hint

The scalar multiplies both components, not only the first.

Next step

Work out $4 \times 2$ and $4 \times (-3)$ .

Full working

4\mathbf{a} = 4(2\mathbf{i} - 3\mathbf{j}) = 8\mathbf{i} - 12\mathbf{j}.

The result is parallel to $\mathbf{a}$ and four times as long.

4. For $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$ and $\mathbf{b} = 3\mathbf{i} - \mathbf{j}$ , find $2\mathbf{a} + 3\mathbf{b}$ .

Hint

Do the two multiplications first, then add. Keep the two answers separate until the end.

Next step

$2\mathbf{a} = 2\mathbf{i} + 4\mathbf{j}$ and $3\mathbf{b} = 9\mathbf{i} - 3\mathbf{j}$ . Now add matching components.

Full working

2\mathbf{a} + 3\mathbf{b} = (2\mathbf{i} + 4\mathbf{j}) + (9\mathbf{i} - 3\mathbf{j}) = 11\mathbf{i} + \mathbf{j}.

5. For $\mathbf{a} = 2\mathbf{i} - \mathbf{j}$ and $\mathbf{b} = \mathbf{i} + 3\mathbf{j}$ , find $3\mathbf{a} - 2\mathbf{b}$ .

Hint

Subtracting $2\mathbf{b}$ means adding $-2\mathbf{b}$ . Change the sign of both components of $2\mathbf{b}$ .

Next step

$3\mathbf{a} = 6\mathbf{i} - 3\mathbf{j}$ and $2\mathbf{b} = 2\mathbf{i} + 6\mathbf{j}$ , so $-2\mathbf{b} = -2\mathbf{i} - 6\mathbf{j}$ .

Full working

3\mathbf{a} - 2\mathbf{b} = (6\mathbf{i} - 3\mathbf{j}) + (-2\mathbf{i} - 6\mathbf{j}) = 4\mathbf{i} - 9\mathbf{j}.

6. Two forces $\mathbf{F}_1 = 5\mathbf{i} + 2\mathbf{j}$ N and $\mathbf{F}_2 = -2\mathbf{i} + 6\mathbf{j}$ N act at a point. Find the resultant, its magnitude and its direction from the positive $x$ -axis.

Hint

This is two jobs. Add the forces to get the resultant, then use Section 10.2 for the magnitude and the direction.

Next step

$\mathbf{F}_1 + \mathbf{F}_2 = 3\mathbf{i} + 8\mathbf{j}$ . Now work out $\sqrt{3^2 + 8^2}$ and $\tan^{-1}\left(\frac{8}{3}\right)$ .

Full working

\mathbf{F}_1 + \mathbf{F}_2 = 3\mathbf{i} + 8\mathbf{j}.

|\mathbf{F}_1 + \mathbf{F}_2| = \sqrt{3^2 + 8^2} = \sqrt{73} \approx 8.54 \text{ N}.

Both components are positive, so the direction is $\tan^{-1}\left(\frac{8}{3}\right) = 69.4^\circ$ to $1$ d.p.

7. Two forces $\mathbf{A} = 4\mathbf{i} - \mathbf{j}$ N and $\mathbf{B} = -\mathbf{i} + 3\mathbf{j}$ N act at a point. Find the force $\mathbf{C}$ that keeps the point still.

Hint

"Keeps the point still" means the three forces are in equilibrium. Their sum must be the zero vector.

Next step

Add $\mathbf{A}$ and $\mathbf{B}$ , then change the sign of the result.

Full working

\mathbf{A} + \mathbf{B} = 3\mathbf{i} + 2\mathbf{j}, \qquad \mathbf{C} = -(3\mathbf{i} + 2\mathbf{j}) = -3\mathbf{i} - 2\mathbf{j} \text{ N}.

8. A ship sails displacement $6\mathbf{i} - 2\mathbf{j}$ km, then $-2\mathbf{i} + 5\mathbf{j}$ km. Find the single displacement from the start, and the distance of the ship from the start.

Hint

Add the two displacements first. The distance is the magnitude of that single displacement.

Next step

The total is $(6 - 2)\mathbf{i} + (-2 + 5)\mathbf{j}$ . Then square and add the components.

Full working

(6\mathbf{i} - 2\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) = 4\mathbf{i} + 3\mathbf{j}.

\text{distance} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ km}.