Column Vectors - TickTockMaths - Free Printable

Let's solve each of the vector operations step by step and express the results in column form. We are given:

$$
\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \quad
\mathbf{b} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}, \quad
\mathbf{c} = \begin{pmatrix} 9 \\ -5 \end{pmatrix}, \quad
\mathbf{d} = \begin{pmatrix} -6 \\ 4 \end{pmatrix}
$$

We will compute each expression one by one.

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1) $\mathbf{a} + \mathbf{b}$

$$
\begin{pmatrix} 1 \\ 2 \end{pmatrix} + \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 + (-2) \\ 2 + 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}
$$

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2) $\mathbf{a} - \mathbf{b}$

$$
\begin{pmatrix} 1 \\ 2 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 - (-2) \\ 2 - 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}
$$

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3) $\mathbf{b} - \mathbf{a}$

$$
\begin{pmatrix} -2 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 - 1 \\ 1 - 2 \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \end{pmatrix}
$$

> Note: This is the negative of (2): $ \mathbf{b} - \mathbf{a} = -(\mathbf{a} - \mathbf{b}) $

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4) $\mathbf{c} - \mathbf{d}$

$$
\begin{pmatrix} 9 \\ -5 \end{pmatrix} - \begin{pmatrix} -6 \\ 4 \end{pmatrix} = \begin{pmatrix} 9 - (-6) \\ -5 - 4 \end{pmatrix} = \begin{pmatrix} 15 \\ -9 \end{pmatrix}
$$

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5) $\mathbf{d} - \mathbf{c}$

$$
\begin{pmatrix} -6 \\ 4 \end{pmatrix} - \begin{pmatrix} 9 \\ -5 \end{pmatrix} = \begin{pmatrix} -6 - 9 \\ 4 - (-5) \end{pmatrix} = \begin{pmatrix} -15 \\ 9 \end{pmatrix}
$$

> Note: This is the negative of (4): $ \mathbf{d} - \mathbf{c} = -(\mathbf{c} - \mathbf{d}) $

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6) $\mathbf{a} - \mathbf{a}$

$$
\begin{pmatrix} 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
$$

> This is the zero vector.

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7) $\mathbf{b} - \mathbf{b}$

$$
\begin{pmatrix} -2 \\ 1 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}
$$

> Also the zero vector.

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8) $\mathbf{a} + \mathbf{b} + \mathbf{c}$

We already have $ \mathbf{a} + \mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix} $

Now add $ \mathbf{c} $:
$$
\begin{pmatrix} -1 \\ 3 \end{pmatrix} + \begin{pmatrix} 9 \\ -5 \end{pmatrix} = \begin{pmatrix} 8 \\ -2 \end{pmatrix}
$$

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9) $\mathbf{a} + \mathbf{b} - \mathbf{c}$

We have $ \mathbf{a} + \mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix} $

Subtract $ \mathbf{c} $:
$$
\begin{pmatrix} -1 \\ 3 \end{pmatrix} - \begin{pmatrix} 9 \\ -5 \end{pmatrix} = \begin{pmatrix} -10 \\ 8 \end{pmatrix}
$$

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10) $\mathbf{a} - \mathbf{b} + \mathbf{c}$

We already have $ \mathbf{a} - \mathbf{b} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} $

Add $ \mathbf{c} $:
$$
\begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 9 \\ -5 \end{pmatrix} = \begin{pmatrix} 12 \\ -4 \end{pmatrix}
$$

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11) $2\mathbf{a} + 2\mathbf{b}$

Factor: $ 2(\mathbf{a} + \mathbf{b}) = 2 \times \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \end{pmatrix} $

Alternatively:
$$
2\mathbf{a} = 2 \times \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}, \quad
2\mathbf{b} = 2 \times \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \end{pmatrix}
$$
Add: $ \begin{pmatrix} 2 - 4 \\ 4 + 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \end{pmatrix} $

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12) $2\mathbf{a} - 2\mathbf{b}$

Factor: $ 2(\mathbf{a} - \mathbf{b}) = 2 \times \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix} $

Or compute directly:
$$
2\mathbf{a} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}, \quad
2\mathbf{b} = \begin{pmatrix} -4 \\ 2 \end{pmatrix}
\Rightarrow \begin{pmatrix} 2 - (-4) \\ 4 - 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}
$$

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13) $2\mathbf{c} - 3\mathbf{d}$

Compute:
$$
2\mathbf{c} = 2 \times \begin{pmatrix} 9 \\ -5 \end{pmatrix} = \begin{pmatrix} 18 \\ -10 \end{pmatrix}, \quad
3\mathbf{d} = 3 \times \begin{pmatrix} -6 \\ 4 \end{pmatrix} = \begin{pmatrix} -18 \\ 12 \end{pmatrix}
$$
Now subtract:
$$
\begin{pmatrix} 18 \\ -10 \end{pmatrix} - \begin{pmatrix} -18 \\ 12 \end{pmatrix} = \begin{pmatrix} 18 + 18 \\ -10 - 12 \end{pmatrix} = \begin{pmatrix} 36 \\ -22 \end{pmatrix}
$$

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14) $4\mathbf{c} - 6\mathbf{d}$

Note: $ 4\mathbf{c} = 2 \times (2\mathbf{c}) $, $ 6\mathbf{d} = 3 \times (2\mathbf{d}) $

But better to compute:
$$
4\mathbf{c} = 4 \times \begin{pmatrix} 9 \\ -5 \end{pmatrix} = \begin{pmatrix} 36 \\ -20 \end{pmatrix}, \quad
6\mathbf{d} = 6 \times \begin{pmatrix} -6 \\ 4 \end{pmatrix} = \begin{pmatrix} -36 \\ 24 \end{pmatrix}
$$
Now subtract:
$$
\begin{pmatrix} 36 \\ -20 \end{pmatrix} - \begin{pmatrix} -36 \\ 24 \end{pmatrix} = \begin{pmatrix} 36 + 36 \\ -20 - 24 \end{pmatrix} = \begin{pmatrix} 72 \\ -44 \end{pmatrix}
$$

> Observe: $ 4\mathbf{c} - 6\mathbf{d} = 2 \times (2\mathbf{c} - 3\mathbf{d}) = 2 \times \begin{pmatrix} 36 \\ -22 \end{pmatrix} = \begin{pmatrix} 72 \\ -44 \end{pmatrix} $

So this is twice question 13.

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15) $20\mathbf{c} - 30\mathbf{d}$

Notice: $ 20\mathbf{c} - 30\mathbf{d} = 10(2\mathbf{c} - 3\mathbf{d}) $

From (13), $ 2\mathbf{c} - 3\mathbf{d} = \begin{pmatrix} 36 \\ -22 \end{pmatrix} $

So:
$$
10 \times \begin{pmatrix} 36 \\ -22 \end{pmatrix} = \begin{pmatrix} 360 \\ -220 \end{pmatrix}
$$

> Also: $ 20\mathbf{c} - 30\mathbf{d} = 5 \times (4\mathbf{c} - 6\mathbf{d}) = 5 \times \begin{pmatrix} 72 \\ -44 \end{pmatrix} = \begin{pmatrix} 360 \\ -220 \end{pmatrix} $

So this is five times question 14, or ten times question 13.

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## ✔ Final Answers (in column form):

| Question | Answer |
|--------|--------|
| 1) $\mathbf{a} + \mathbf{b}$ | $\begin{pmatrix} -1 \\ 3 \end{pmatrix}$ |
| 2) $\mathbf{a} - \mathbf{b}$ | $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ |
| 3) $\mathbf{b} - \mathbf{a}$ | $\begin{pmatrix} -3 \\ -1 \end{pmatrix}$ |
| 4) $\mathbf{c} - \mathbf{d}$ | $\begin{pmatrix} 15 \\ -9 \end{pmatrix}$ |
| 5) $\mathbf{d} - \mathbf{c}$ | $\begin{pmatrix} -15 \\ 9 \end{pmatrix}$ |
| 6) $\mathbf{a} - \mathbf{a}$ | $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ |
| 7) $\mathbf{b} - \mathbf{b}$ | $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ |
| 8) $\mathbf{a} + \mathbf{b} + \mathbf{c}$ | $\begin{pmatrix} 8 \\ -2 \end{pmatrix}$ |
| 9) $\mathbf{a} + \mathbf{b} - \mathbf{c}$ | $\begin{pmatrix} -10 \\ 8 \end{pmatrix}$ |
| 10) $\mathbf{a} - \mathbf{b} + \mathbf{c}$ | $\begin{pmatrix} 12 \\ -4 \end{pmatrix}$ |
| 11) $2\mathbf{a} + 2\mathbf{b}$ | $\begin{pmatrix} -2 \\ 6 \end{pmatrix}$ |
| 12) $2\mathbf{a} - 2\mathbf{b}$ | $\begin{pmatrix} 6 \\ 2 \end{pmatrix}$ |
| 13) $2\mathbf{c} - 3\mathbf{d}$ | $\begin{pmatrix} 36 \\ -22 \end{pmatrix}$ |
| 14) $4\mathbf{c} - 6\mathbf{d}$ | $\begin{pmatrix} 72 \\ -44 \end{pmatrix}$ |
| 15) $20\mathbf{c} - 30\mathbf{d}$ | $\begin{pmatrix} 360 \\ -220 \end{pmatrix}$ |

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## 🔍 Links Between Questions:

1. (3) = -(2) → $ \mathbf{b} - \mathbf{a} = -(\mathbf{a} - \mathbf{b}) $
2. (5) = -(4) → $ \mathbf{d} - \mathbf{c} = -(\mathbf{c} - \mathbf{d}) $
3. (6) and (7) → Both are zero vectors.
4. (11) = 2 × (1) → $ 2(\mathbf{a} + \mathbf{b}) $
5. (12) = 2 × (2) → $ 2(\mathbf{a} - \mathbf{b}) $
6. (14) = 2 × (13) → $ 4\mathbf{c} - 6\mathbf{d} = 2(2\mathbf{c} - 3\mathbf{d}) $
7. (15) = 10 × (13) or 5 × (14) → Scalar multiples

✔ These show that many questions are scalar multiples or negatives of earlier ones. This highlights vector addition properties, scalar multiplication, and commutative/distributive laws.

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