10.1 Vectors in two and three dimensions — Practise with help

Work each one first, closed book. Before you write a vector, decide which point you are moving from and which point you are moving to.

Warm-up: remember

Work out $\sqrt{3^2 + 4^2}$ .
Work out $5 - (-3)$ and $-2 - 6$ .
The point $P$ is $3$ units right and $4$ units up from the origin $O$ . Write the coordinates of $P$ .

Answers

$3^2 + 4^2 = 9 + 16 = 25$ , so $\sqrt{25} = 5$ .
$5 - (-3) = 5 + 3 = 8$ , and $-2 - 6 = -8$ .
Moving right adds to $x$ and moving up adds to $y$ , so $P$ is $(3, 4)$ .

If question 2 was hard, revisit subtracting negative numbers before you continue.

Practise with help

1. Write the displacement from $A(2, 1)$ to $B(7, 5)$ as a column vector and in $\mathbf{i}$ , $\mathbf{j}$ form.

Hint

Subtract the coordinates of $A$ from the coordinates of $B$ . Keep the $x$ values on the top.

Next step

The $x$ -component is $7 - 2 = 5$ . The $y$ -component is $5 - 1 = 4$ .

Full working

\vec{AB} = \begin{pmatrix} 7 - 2 \\ 5 - 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix} = 5\mathbf{i} + 4\mathbf{j}.

The vector moves $5$ units right and $4$ units up.

2. Write the displacement from $B(7, 5)$ to $A(2, 1)$ .

Hint

This displacement is the reverse of the one from question 1. Predict the signs before you calculate.

Next step

Subtract $B$ from $A$ : $2 - 7$ on the top and $1 - 5$ on the bottom.

Full working

\vec{BA} = \begin{pmatrix} 2 - 7 \\ 1 - 5 \end{pmatrix} = \begin{pmatrix} -5 \\ -4 \end{pmatrix} = -5\mathbf{i} - 4\mathbf{j}.

This is $-\vec{AB}$ , so the arrow points the opposite way.

3. A vector is $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$ . Write down $3\mathbf{a}$ and $-\mathbf{a}$ .

Hint

Multiplying by a scalar multiplies each component separately. A negative scalar also reverses the direction.

Next step

For $3\mathbf{a}$ , work out $3 \times 2$ and $3 \times 3$ . For $-\mathbf{a}$ , change the sign of each component.

Full working

3\mathbf{a} = 6\mathbf{i} + 9\mathbf{j}, \qquad -\mathbf{a} = -2\mathbf{i} - 3\mathbf{j}.

Both new vectors are parallel to $\mathbf{a}$ .

4. Show that $\mathbf{u} = \mathbf{i} - 2\mathbf{j}$ and $\mathbf{v} = -3\mathbf{i} + 6\mathbf{j}$ are parallel.

Hint

Compare the components. Work out what number multiplies each component of $\mathbf{u}$ to give the matching component of $\mathbf{v}$ .

Next step

The $x$ -component changes from $1$ to $-3$ , which suggests a multiplier of $-3$ . Check the $y$ -component too.

Full working

-3\mathbf{u} = -3(\mathbf{i} - 2\mathbf{j}) = -3\mathbf{i} + 6\mathbf{j} = \mathbf{v}.

The two vectors are parallel. The multiplier is negative, so they point in opposite directions.

5. Write the displacement from $P(1, 0, 2)$ to $Q(4, 3, -2)$ in $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ form.

Hint

This time there are three coordinates. Subtract $P$ from $Q$ in each position: $x$ , then $y$ , then $z$ .

Next step

The components are $4 - 1$ , $3 - 0$ and $-2 - 2$ . Take care with the last one.

Full working

\vec{PQ} = \begin{pmatrix} 4 - 1 \\ 3 - 0 \\ -2 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix} = 3\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}.

The negative $z$ -component means the displacement is downwards.

6. A vector is $\mathbf{w} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ . Write down the column form of $\mathbf{w}$ and of $\frac{1}{2}\mathbf{w}$ .

Hint

The column has three rows, one for each of $x$ , $y$ and $z$ . Halving works on each row.

Next step

Read the coefficients: $4$ for $\mathbf{i}$ , $-1$ for $\mathbf{j}$ and $2$ for $\mathbf{k}$ . Then halve each one.

Full working

\mathbf{w} = \begin{pmatrix} 4 \\ -1 \\ 2 \end{pmatrix}, \qquad \frac{1}{2}\mathbf{w} = \begin{pmatrix} 2 \\ -\frac{1}{2} \\ 1 \end{pmatrix}.

The two vectors are parallel, and $\frac{1}{2}\mathbf{w}$ points the same way.

7. The points $A(1, 2)$ , $B(4, 6)$ and $C(7, 10)$ lie on a straight line. Show that $\vec{AB}$ and $\vec{BC}$ are equal.

Hint

Work out each displacement separately, then compare the two columns.

Next step

$\vec{AB}$ has components $4 - 1$ and $6 - 2$ . For $\vec{BC}$ , subtract $B$ from $C$ .

Full working

\vec{AB} = \begin{pmatrix} 4 - 1 \\ 6 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}, \qquad \vec{BC} = \begin{pmatrix} 7 - 4 \\ 10 - 6 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}.

The two displacements are equal. Equal displacement vectors in the same straight line mean the same step repeats, so $A$ , $B$ and $C$ stay on one line.