10.3 Vector addition and scalar multiplication — Practise

Work through these on your own. For each question, decide whether the answer is a vector, a magnitude or a direction, and finish the calculation in that form. Add and scale the components before you take any square root.

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

Add $\mathbf{a} = 2\mathbf{i} + 5\mathbf{j}$ and $\mathbf{b} = 3\mathbf{i} - 4\mathbf{j}$ .
Add $\mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}$ .
Multiply $\mathbf{a} = 2\mathbf{i} - 5\mathbf{j}$ by $-3$ .
For $\mathbf{a} = 3\mathbf{i} - \mathbf{j}$ and $\mathbf{b} = \mathbf{i} + 4\mathbf{j}$ , find $2\mathbf{a} + 3\mathbf{b}$ .
For $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$ and $\mathbf{b} = 2\mathbf{i} - 3\mathbf{j}$ , find $4\mathbf{a} - 2\mathbf{b}$ .
Two forces $\mathbf{F}_1 = 3\mathbf{i} + 4\mathbf{j}$ N and $\mathbf{F}_2 = 4\mathbf{i} - 3\mathbf{j}$ N act at a point. Find the resultant force, its magnitude and its direction from the positive $x$ -axis.
Forces $\mathbf{A} = 5\mathbf{i} - 2\mathbf{j}$ N and $\mathbf{B} = -3\mathbf{i} + 7\mathbf{j}$ N act at a point. Find the force $\mathbf{C}$ that keeps the point still.
Show that $\mathbf{a} = 2\mathbf{i} - \mathbf{j}$ , $\mathbf{b} = -\mathbf{i} + 4\mathbf{j}$ and $\mathbf{c} = -\mathbf{i} - 3\mathbf{j}$ are in equilibrium.
A ship has displacements $3\mathbf{i} - 4\mathbf{j}$ km, $-\mathbf{i} + 2\mathbf{j}$ km and $2\mathbf{i} + \mathbf{j}$ km. Find the single displacement from the start and the distance from the start.
For $\mathbf{a} = 2\mathbf{i} - \mathbf{j}$ and $\mathbf{b} = -\mathbf{i} + 2\mathbf{j}$ , find $3\mathbf{a} - 4\mathbf{b}$ .
A boat can move at $3\mathbf{i} + 4\mathbf{j}$ m/s in still water. A current of $-\mathbf{i} + 2\mathbf{j}$ m/s flows. Find the boat's resultant velocity, its speed and its direction from the positive $x$ -axis.
A drone flies displacement $6\mathbf{i} + \mathbf{j}$ m, then $-4\mathbf{i} + 3\mathbf{j}$ m, then $\mathbf{i} - \mathbf{j}$ m. Find its single displacement and its distance from the start.

Answers and concise solutions

1. The sum

Answer: Add the matching components: $\mathbf{a} + \mathbf{b} = (2 + 3)\mathbf{i} + (5 - 4)\mathbf{j} = 5\mathbf{i} + \mathbf{j}$ . The resultant is the column $\begin{pmatrix} 5 \\ 1 \end{pmatrix}$ .

2. The sum

Answer: Add each row: $(1 + 2)\mathbf{i} + (-2 + 2)\mathbf{j} + (3 - 5)\mathbf{k} = 3\mathbf{i} - 2\mathbf{k}$ . The $\mathbf{j}$ row cancels to $0$ , so there is no $\mathbf{j}$ term.

3. The vector $-3\mathbf{a}$

Answer: $-3\mathbf{a} = -3(2\mathbf{i} - 5\mathbf{j}) = -6\mathbf{i} + 15\mathbf{j}$ . The negative scalar reverses the direction, and the new vector is three times as long as $\mathbf{a}$ .

4. The vector

Answer: $2\mathbf{a} = 6\mathbf{i} - 2\mathbf{j}$ and $3\mathbf{b} = 3\mathbf{i} + 12\mathbf{j}$ . Adding: $2\mathbf{a} + 3\mathbf{b} = 9\mathbf{i} + 10\mathbf{j}$ .

5. The vector

Answer: $4\mathbf{a} = 4\mathbf{i} + 8\mathbf{j}$ and $2\mathbf{b} = 4\mathbf{i} - 6\mathbf{j}$ . Subtracting: $4\mathbf{a} - 2\mathbf{b} = (4 - 4)\mathbf{i} + (8 - (-6))\mathbf{j} = 14\mathbf{j}$ .

6. The resultant, its magnitude and direction

Answer: The resultant is $(3 + 4)\mathbf{i} + (4 - 3)\mathbf{j} = 7\mathbf{i} + \mathbf{j}$ . Its magnitude is $\sqrt{7^2 + 1^2} = \sqrt{50} = 5\sqrt{2} \approx 7.07$ N, and its direction is $\tan^{-1}\left(\dfrac{1}{7}\right) = 8.1^\circ$ to $1$ d.p.

7. The third force

Answer: For the point to stay still, $\mathbf{A} + \mathbf{B} + \mathbf{C} = \mathbf{0}$ . Now $\mathbf{A} + \mathbf{B} = 2\mathbf{i} + 5\mathbf{j}$ N, so $\mathbf{C} = -2\mathbf{i} - 5\mathbf{j}$ N.

8. The total

Answer: Add the three vectors: $(2 - 1 - 1)\mathbf{i} + (-1 + 4 - 3)\mathbf{j} = 0\mathbf{i} + 0\mathbf{j}$ . The total is the zero vector, so the three vectors are in equilibrium.

9. The displacement and the distance

Answer: The total displacement is $(3 - 1 + 2)\mathbf{i} + (-4 + 2 + 1)\mathbf{j} = 4\mathbf{i} - \mathbf{j}$ km. The distance from the start is $\sqrt{4^2 + (-1)^2} = \sqrt{17}$ km.

10. The vector

Answer: $3\mathbf{a} = 6\mathbf{i} - 3\mathbf{j}$ and $4\mathbf{b} = -4\mathbf{i} + 8\mathbf{j}$ . Subtracting: $3\mathbf{a} - 4\mathbf{b} = (6 - (-4))\mathbf{i} + (-3 - 8)\mathbf{j} = 10\mathbf{i} - 11\mathbf{j}$ .

11. The velocity, the speed and the direction

Answer: The resultant velocity is $(3 - 1)\mathbf{i} + (4 + 2)\mathbf{j} = 2\mathbf{i} + 6\mathbf{j}$ m/s. Its speed is $\sqrt{2^2 + 6^2} = \sqrt{40} = 2\sqrt{10} \approx 6.32$ m/s. Both components are positive, so the direction is $\tan^{-1}\left(\dfrac{6}{2}\right) = 71.6^\circ$ to $1$ d.p.

12. The displacement and the distance

Answer: The total displacement is $(6 - 4 + 1)\mathbf{i} + (1 + 3 - 1)\mathbf{j} = 3\mathbf{i} + 3\mathbf{j}$ m. The distance from the start is $\sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \approx 4.24$ m.