GCSE Maths Tutor worksheet SS-VS-01 on vectors, including exercises on vector operations and geometric interpretations.

Let's solve each question step by step from the vector worksheet you've provided.

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Question 1:

Given:
$$
x = \begin{pmatrix} 3 \\ 5 \end{pmatrix},\quad y = \begin{pmatrix} 2 \\ -4 \end{pmatrix},\quad z = \begin{pmatrix} 3 \\ 0 \end{pmatrix}
$$

We perform vector addition and subtraction component-wise.

(a) $ x + y $
$$
x + y = \begin{pmatrix} 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 3+2 \\ 5+(-4) \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}
$$

(b) $ y + z $
$$
y + z = \begin{pmatrix} 2 \\ -4 \end{pmatrix} + \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}
$$

(c) $ x + z $
$$
x + z = \begin{pmatrix} 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}
$$

(d) $ x - y $
$$
x - y = \begin{pmatrix} 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 3-2 \\ 5-(-4) \end{pmatrix} = \begin{pmatrix} 1 \\ 9 \end{pmatrix}
$$

(e) $ y - z $
$$
y - z = \begin{pmatrix} 2 \\ -4 \end{pmatrix} - \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ -4 \end{pmatrix}
$$

(f) $ x - z $
$$
x - z = \begin{pmatrix} 3 \\ 5 \end{pmatrix} - \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}
$$

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Question 2:

Given:
$$
p = \begin{pmatrix} 1 \\ 3 \end{pmatrix},\quad q = \begin{pmatrix} 2 \\ -3 \end{pmatrix},\quad r = \begin{pmatrix} 3 \\ -2 \end{pmatrix}
$$

Solve for $ x $ in each equation.

(a) $ x = p - q $
$$
x = \begin{pmatrix} 1 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 1-2 \\ 3-(-3) \end{pmatrix} = \begin{pmatrix} -1 \\ 6 \end{pmatrix}
$$

(b) $ x - q = p $ → $ x = p + q $
$$
x = \begin{pmatrix} 1 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix}
$$

(c) $ q = r - x $ → $ x = r - q $
$$
x = \begin{pmatrix} 3 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}
$$

(d) $ r + p = 2x $ → $ x = \frac{r + p}{2} $
$$
r + p = \begin{pmatrix} 3 \\ -2 \end{pmatrix} + \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}
\Rightarrow x = \frac{1}{2} \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0.5 \end{pmatrix}
$$

(e) $ 3r = 2q - x $ → Solve for $ x $:
$$
x = 2q - 3r
$$
First compute:
$$
2q = 2 \cdot \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ -6 \end{pmatrix},\quad
3r = 3 \cdot \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}
$$
$$
x = \begin{pmatrix} 4 \\ -6 \end{pmatrix} - \begin{pmatrix} 9 \\ -6 \end{pmatrix} = \begin{pmatrix} -5 \\ 0 \end{pmatrix}
$$

(f) $ p = 3r - 2x $ → Solve for $ x $:
$$
2x = 3r - p \Rightarrow x = \frac{3r - p}{2}
$$
Compute:
$$
3r = \begin{pmatrix} 9 \\ -6 \end{pmatrix},\quad p = \begin{pmatrix} 1 \\ 3 \end{pmatrix}
\Rightarrow 3r - p = \begin{pmatrix} 8 \\ -9 \end{pmatrix}
\Rightarrow x = \frac{1}{2} \begin{pmatrix} 8 \\ -9 \end{pmatrix} = \begin{pmatrix} 4 \\ -4.5 \end{pmatrix}
$$

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Question 3: Find the magnitude of each column vector (to 2 d.p.)

The magnitude of a vector $ \begin{pmatrix} a \\ b \end{pmatrix} $ is $ \sqrt{a^2 + b^2} $

(a) $ \begin{pmatrix} 2 \\ 1 \end{pmatrix} $
$$
|v| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.24
$$

(b) $ \begin{pmatrix} -3 \\ 2 \end{pmatrix} $
$$
|v| = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \approx 3.61
$$

(c) $ \begin{pmatrix} 5 \\ -1 \end{pmatrix} $
$$
|v| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.10
$$

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Question 4: Angle with positive x-axis (anti-clockwise)

Use $ \theta = \tan^{-1}\left(\frac{y}{x}\right) $, but consider quadrant.

(a) $ \begin{pmatrix} 1 \\ 1 \end{pmatrix} $
This is in first quadrant:
$$
\theta = \tan^{-1}(1/1) = \tan^{-1}(1) = 45^\circ
$$

(b) $ \begin{pmatrix} -1 \\ 1 \end{pmatrix} $
Second quadrant:
$$
\tan^{-1}(1/-1) = \tan^{-1}(-1) = -45^\circ, \text{ but add } 180^\circ \Rightarrow 135^\circ
$$

(c) $ \begin{pmatrix} -1 \\ -1 \end{pmatrix} $
Third quadrant:
$$
\tan^{-1}((-1)/(-1)) = \tan^{-1}(1) = 45^\circ, \text{ but add } 180^\circ \Rightarrow 225^\circ
$$

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✔ Final Answers:

#### Question 1:
(a) $ \begin{pmatrix} 5 \\ 1 \end{pmatrix} $
(b) $ \begin{pmatrix} 5 \\ -4 \end{pmatrix} $
(c) $ \begin{pmatrix} 6 \\ 5 \end{pmatrix} $
(d) $ \begin{pmatrix} 1 \\ 9 \end{pmatrix} $
(e) $ \begin{pmatrix} -1 \\ -4 \end{pmatrix} $
(f) $ \begin{pmatrix} 0 \\ 5 \end{pmatrix} $

#### Question 2:
(a) $ \begin{pmatrix} -1 \\ 6 \end{pmatrix} $
(b) $ \begin{pmatrix} 3 \\ 0 \end{pmatrix} $
(c) $ \begin{pmatrix} 1 \\ 1 \end{pmatrix} $
(d) $ \begin{pmatrix} 2 \\ 0.5 \end{pmatrix} $
(e) $ \begin{pmatrix} -5 \\ 0 \end{pmatrix} $
(f) $ \begin{pmatrix} 4 \\ -4.5 \end{pmatrix} $

#### Question 3: Magnitudes
(a) $ 2.24 $
(b) $ 3.61 $
(c) $ 5.10 $

#### Question 4: Angles
(a) $ 45^\circ $
(b) $ 135^\circ $
(c) $ 225^\circ $

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