This worksheet helps students practice calculating vector magnitudes using grids and column notation.

To solve the problems in the worksheet, we need to determine the column vectors and their magnitudes for each given vector. Let's go through each part step by step.

Section A: Draw and write the vector, and state its magnitude as a surd where appropriate.

#### 1) Vector \( \mathbf{a} \)

- Column Vector: From the grid, vector \( \mathbf{a} \) moves 3 units right and 2 units up.
\[
\mathbf{a} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}
\]

- Magnitude:
\[
|\mathbf{a}| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}
\]

#### 2) Vector \( \mathbf{b} \)

- Column Vector: From the grid, vector \( \mathbf{b} \) moves 2 units left and 4 units down.
\[
\mathbf{b} = \begin{pmatrix} -2 \\ -4 \end{pmatrix}
\]

- Magnitude:
\[
|\mathbf{b}| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}
\]

#### 3) Vector \( \mathbf{a} + \mathbf{b} \)

- Column Vector:
\[
\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} + \begin{pmatrix} -2 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 + (-2) \\ 2 + (-4) \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}
\]

- Magnitude:
\[
|\mathbf{a} + \mathbf{b}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}
\]

#### 4) Vector \( \mathbf{a} - \mathbf{b} \)

- Column Vector:
\[
\mathbf{a} - \mathbf{b} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} - \begin{pmatrix} -2 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 - (-2) \\ 2 - (-4) \end{pmatrix} = \begin{pmatrix} 3 + 2 \\ 2 + 4 \end{pmatrix} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}
\]

- Magnitude:
\[
|\mathbf{a} - \mathbf{b}| = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}
\]

#### 5) Vector \( 2\mathbf{a} \)

- Column Vector:
\[
2\mathbf{a} = 2 \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \cdot 3 \\ 2 \cdot 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}
\]

- Magnitude:
\[
|2\mathbf{a}| = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
\]

#### 6) Vector \( \mathbf{a} - \frac{1}{2}\mathbf{b} \)

- Column Vector:
\[
\frac{1}{2}\mathbf{b} = \frac{1}{2} \begin{pmatrix} -2 \\ -4 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \cdot (-2) \\ \frac{1}{2} \cdot (-4) \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \end{pmatrix}
\]
\[
\mathbf{a} - \frac{1}{2}\mathbf{b} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} - \begin{pmatrix} -1 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 - (-1) \\ 2 - (-2) \end{pmatrix} = \begin{pmatrix} 3 + 1 \\ 2 + 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}
\]

- Magnitude:
\[
|\mathbf{a} - \frac{1}{2}\mathbf{b}| = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

#### 7) Vector \( \mathbf{c} \)

- Column Vector: Given \( |\mathbf{c}| = 2\sqrt{5} \) and the second component is \(-2\). Let the first component be \( x \).
\[
|\mathbf{c}| = \sqrt{x^2 + (-2)^2} = 2\sqrt{5}
\]
\[
\sqrt{x^2 + 4} = 2\sqrt{5}
\]
Squaring both sides:
\[
x^2 + 4 = 20
\]
\[
x^2 = 16
\]
\[
x = \pm 4
\]
So, the column vector can be:
\[
\mathbf{c} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} \quad \text{or} \quad \mathbf{c} = \begin{pmatrix} -4 \\ -2 \end{pmatrix}
\]

#### 8) Vector with Magnitude \( \sqrt{34} \)

- Column Vector: Given the second component is \(-5\) and the magnitude is \( \sqrt{34} \). Let the first component be \( y \).
\[
|\mathbf{v}| = \sqrt{y^2 + (-5)^2} = \sqrt{34}
\]
\[
\sqrt{y^2 + 25} = \sqrt{34}
\]
Squaring both sides:
\[
y^2 + 25 = 34
\]
\[
y^2 = 9
\]
\[
y = \pm 3
\]
So, the column vector can be:
\[
\mathbf{v} = \begin{pmatrix} 3 \\ -5 \end{pmatrix} \quad \text{or} \quad \mathbf{v} = \begin{pmatrix} -3 \\ -5 \end{pmatrix}
\]

Final Answers

\[
\boxed{
\begin{array}{ll}
1) & \mathbf{a} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}, \quad |\mathbf{a}| = \sqrt{13} \\
2) & \mathbf{b} = \begin{pmatrix} -2 \\ -4 \end{pmatrix}, \quad |\mathbf{b}| = 2\sqrt{5} \\
3) & \mathbf{a} + \mathbf{b} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}, \quad |\mathbf{a} + \mathbf{b}| = \sqrt{5} \\
4) & \mathbf{a} - \mathbf{b} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}, \quad |\mathbf{a} - \mathbf{b}| = \sqrt{61} \\
5) & 2\mathbf{a} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}, \quad |2\mathbf{a}| = 2\sqrt{13} \\
6) & \mathbf{a} - \frac{1}{2}\mathbf{b} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}, \quad |\mathbf{a} - \frac{1}{2}\mathbf{b}| = 4\sqrt{2} \\
7) & \mathbf{c} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} \text{ or } \begin{pmatrix} -4 \\ -2 \end{pmatrix}, \quad |\mathbf{c}| = 2\sqrt{5} \\
8) & \mathbf{v} = \begin{pmatrix} 3 \\ -5 \end{pmatrix} \text{ or } \begin{pmatrix} -3 \\ -5 \end{pmatrix}, \quad |\mathbf{v}| = \sqrt{34}
\end{array}
}
\]