10.1 Vectors in two and three dimensions — Practise

Work through these on your own. Before you write a vector, name the point you start from and the point you finish at. Check the order of the components in every column.

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

Write the displacement from $A(1, 2)$ to $B(5, 7)$ as a column vector, and in $\mathbf{i}$ , $\mathbf{j}$ form.
Write the displacement from $B(5, 7)$ to $A(1, 2)$ as a column vector.
A vector is $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$ . Write down $3\mathbf{a}$ in column form.
A vector is $\mathbf{w} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ . Write down $-\mathbf{w}$ in column form.
Show that $\mathbf{v} = 6\mathbf{i} - 3\mathbf{j}$ is parallel to $\mathbf{u} = 2\mathbf{i} - \mathbf{j}$ , and state the multiplier.
The points are $P(1, 5)$ and $Q(4, 3)$ . Show that $\vec{PQ}$ equals $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j}$ .
Write the displacement from $P(1, 0, 2)$ to $Q(4, 3, -2)$ in column form.
A vector is $\mathbf{w} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ . Write down $\frac{1}{2}\mathbf{w}$ in column form.
Write $6\mathbf{i} - 5\mathbf{j}$ as a column vector. State the top number and the bottom number.
The points are $A(2, 3)$ , $B(5, 8)$ and $C(8, 13)$ . Show that $\vec{AB}$ and $\vec{BC}$ are equal.
Show that $\mathbf{b} = -4\mathbf{i} - 6\mathbf{j}$ is parallel to $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$ , and state the multiplier.
A displacement is $\vec{AB} = 3\mathbf{i} - 4\mathbf{j}$ . Write down $\vec{BA}$ and $2\vec{AB}$ .

Answers and concise solutions

1. The displacement

Answer: Subtract the coordinates of $A$ from those of $B$ : $\vec{AB} = \begin{pmatrix} 5 - 1 \\ 7 - 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \end{pmatrix} = 4\mathbf{i} + 5\mathbf{j}$ . The vector moves $4$ units east and $5$ units north.

2. The displacement

Answer: Subtract the coordinates of $B$ from those of $A$ : $\vec{BA} = \begin{pmatrix} 1 - 5 \\ 2 - 7 \end{pmatrix} = \begin{pmatrix} -4 \\ -5 \end{pmatrix}$ . This is $-\vec{AB}$ , so it points the opposite way.

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

Answer: Multiplying by a scalar multiplies each component: $3\mathbf{a} = 3(2\mathbf{i} + 3\mathbf{j}) = 6\mathbf{i} + 9\mathbf{j}$ , which is the column $\begin{pmatrix} 6 \\ 9 \end{pmatrix}$ . The new vector is parallel to $\mathbf{a}$ and three times as long.

4. The vector $-\mathbf{w}$

Answer: Multiplying by $-1$ changes the sign of every component: $-\mathbf{w} = -4\mathbf{i} + \mathbf{j} - 2\mathbf{k} = \begin{pmatrix} -4 \\ 1 \\ -2 \end{pmatrix}$ . The vector is parallel to $\mathbf{w}$ and points the opposite way.

5. The multiplier $k$

Answer: Each component of $\mathbf{v}$ is $3$ times the matching component of $\mathbf{u}$ : $\mathbf{v} = 3\mathbf{u}$ , so $k = 3$ . The multiplier is positive, so the two vectors point the same way.

6. The displacement $\vec{PQ}$

Answer: $\vec{PQ} = \begin{pmatrix} 4 - 1 \\ 3 - 5 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} = 3\mathbf{i} - 2\mathbf{j}$ . Both components match $\mathbf{a}$ , so the vectors are equal.

7. The displacement

Answer: Subtract each coordinate of $P$ from the matching coordinate of $Q$ : $\vec{PQ} = \begin{pmatrix} 4 - 1 \\ 3 - 0 \\ -2 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix}$ . The negative $z$ -component means the displacement is downwards.

8. The vector $\frac{1}{2}\mathbf{w}$

Answer: Halve each component: $\frac{1}{2}\mathbf{w} = 2\mathbf{i} - \frac{1}{2}\mathbf{j} + \mathbf{k} = \begin{pmatrix} 2 \\ -\frac{1}{2} \\ 1 \end{pmatrix}$ . The two vectors are parallel, and $\frac{1}{2}\mathbf{w}$ points the same way as $\mathbf{w}$ .

9. The two components

Answer: The coefficient of $\mathbf{i}$ is the $x$ -component, which goes on top. The coefficient of $\mathbf{j}$ is the $y$ -component, which goes below. The column is $\begin{pmatrix} 6 \\ -5 \end{pmatrix}$ .

10. Both displacements

Answer: $\vec{AB} = \begin{pmatrix} 5 - 2 \\ 8 - 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$ and $\vec{BC} = \begin{pmatrix} 8 - 5 \\ 13 - 8 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \end{pmatrix}$ . The two displacements are equal, so the same step repeats and the three points lie on one straight line.

11. The multiplier $k$

Answer: Both components divide by the same number: $\mathbf{b} = -2\mathbf{a}$ , so $k = -2$ . The multiplier is negative, so $\mathbf{b}$ is parallel to $\mathbf{a}$ but points the opposite way.

12. Both vectors

Answer: Changing the order reverses the signs: $\vec{BA} = -3\mathbf{i} + 4\mathbf{j}$ . Doubling multiplies each component by $2$ : $2\vec{AB} = 6\mathbf{i} - 8\mathbf{j}$ . The first points the opposite way; the second points the same way and is twice as long.