Vectors in two dimensions

1. Overview

Vectors in two dimensions are fundamental to Additional Mathematics, providing a way to represent quantities with both magnitude and direction. This topic is crucial for solving problems involving geometry, motion, and forces. Mastering vectors allows you to model real-world scenarios, such as calculating the resultant force on an object or determining the position of a moving particle. Expect vector questions in both Paper 1 (non-calculator) and Paper 2, often integrated with geometry or kinematics. A strong understanding of vector notation, magnitude, and operations is essential for success in the 0606 examination.

Key Definitions

• Scalar: A quantity with magnitude only (e.g., speed, time, distance).
• Vector: A quantity with both magnitude and direction (e.g., velocity, displacement, force).
• Position Vector: A vector that represents the position of a point relative to a fixed origin $O$, usually denoted as $\vec{OA}$ or $\mathbf{a}$.
• Magnitude: The length of a vector, denoted by $|\mathbf{a}|$ or $|\vec{AB}|$.
• Unit Vector: A vector with a magnitude of exactly 1 unit, denoted as $\mathbf{\hat{a}}$.
• Resultant Vector: The vector produced by adding two or more vectors together.
• Collinear: Points that lie on the same straight line; vectors are collinear if one is a scalar multiple of the other.

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Core Content

A. Vector Notation

Vectors can be written in three main ways. You must be comfortable switching between them:

1. Column Vectors: $\begin{pmatrix} x \ y \end{pmatrix}$
2. Unit Vector Form: $x\mathbf{i} + y\mathbf{j}$, where $\mathbf{i}$ is the unit vector $\begin{pmatrix} 1 \ 0 \end{pmatrix}$ and $\mathbf{j}$ is $\begin{pmatrix} 0 \ 1 \end{pmatrix}$.
3. Geometric Notation: $\vec{AB}$ represents the vector from point $A$ to point $B$. Note that $\vec{BA} = -\vec{AB}$.

B. Magnitude and Unit Vectors

To find the magnitude of $\mathbf{a} = \begin{pmatrix} x \ y \end{pmatrix}$: $$|\mathbf{a}| = \sqrt{x^2 + y^2}$$ Always leave your answer in exact surd form unless specified otherwise.

A unit vector in the direction of $\mathbf{a}$ is found by dividing the vector by its magnitude: $$\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$$

Worked example 1 — Magnitude and Unit Vector

Question: Given $\mathbf{p} = 5\mathbf{i} - 12\mathbf{j}$, find $|\mathbf{p}|$ and the unit vector in the direction of $\mathbf{p}$.

1. Find the magnitude of $\mathbf{p}$: $|\mathbf{p}| = \sqrt{5^2 + (-12)^2}$ (Apply the magnitude formula)

2. Simplify: $|\mathbf{p}| = \sqrt{25 + 144}$ (Evaluate the squares)

3. Further simplification: $|\mathbf{p}| = \sqrt{169}$ (Add the terms)

4. Calculate the square root: $|\mathbf{p}| = 13$ (Magnitude is a scalar, so it's just a number)

5. State the magnitude: $|\mathbf{p}| = 13$

6. Find the unit vector $\mathbf{\hat{p}}$: $\mathbf{\hat{p}} = \frac{\mathbf{p}}{|\mathbf{p}|}$ (Apply the unit vector formula)

7. Substitute the values: $\mathbf{\hat{p}} = \frac{1}{13}(5\mathbf{i} - 12\mathbf{j})$ (Substitute magnitude and vector)

8. Distribute the scalar: $\mathbf{\hat{p}} = \frac{5}{13}\mathbf{i} - \frac{12}{13}\mathbf{j}$ (Multiply each component by the scalar)

9. State the unit vector: $\mathbf{\hat{p}} = \frac{5}{13}\mathbf{i} - \frac{12}{13}\mathbf{j}$

Answer: $|\mathbf{p}| = \mathbf{13}$, $\mathbf{\hat{p}} = \frac{5}{13}\mathbf{i} - \frac{12}{13}\mathbf{j}$

C. Vector Geometry and Ratios

To find the vector between two points $A$ and $B$ given their position vectors $\mathbf{a}$ and $\mathbf{b}$: $$\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$$

Worked Example 2 — Vector Ratio Problem

Question: Relative to an origin $O$, the position vector of $A$ is $2\mathbf{a}$ and $B$ is $5\mathbf{b}$. The point $P$ lies on $AB$ such that $AP:PB = 2:1$. Find $\vec{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

1. Find $\vec{AB}$: $\vec{AB} = \vec{OB} - \vec{OA}$ (Vector between two points)
2. Substitute the position vectors: $\vec{AB} = 5\mathbf{b} - 2\mathbf{a}$ (Substitute $\vec{OB} = 5\mathbf{b}$ and $\vec{OA} = 2\mathbf{a}$)
3. Express $\vec{AP}$ as a fraction of $\vec{AB}$: $\vec{AP} = \frac{2}{3}\vec{AB}$ (Since $AP:PB = 2:1$, $AP$ is $\frac{2}{2+1} = \frac{2}{3}$ of $AB$)
4. Calculate $\vec{AP}$: $\vec{AP} = \frac{2}{3}(5\mathbf{b} - 2\mathbf{a})$ (Substitute the expression for $\vec{AB}$)
5. Expand: $\vec{AP} = \frac{10}{3}\mathbf{b} - \frac{4}{3}\mathbf{a}$ (Distribute the scalar $\frac{2}{3}$)
6. Find $\vec{OP}$ using the path $O \to A \to P$: $\vec{OP} = \vec{OA} + \vec{AP}$ (Vector addition)
7. Substitute the vectors: $\vec{OP} = 2\mathbf{a} + (\frac{10}{3}\mathbf{b} - \frac{4}{3}\mathbf{a})$ (Substitute $\vec{OA} = 2\mathbf{a}$ and the expression for $\vec{AP}$)
8. Simplify: $\vec{OP} = \frac{6}{3}\mathbf{a} - \frac{4}{3}\mathbf{a} + \frac{10}{3}\mathbf{b}$ (Rewrite $2\mathbf{a}$ as $\frac{6}{3}\mathbf{a}$)
9. Combine like terms: $\vec{OP} = \frac{2}{3}\mathbf{a} + \frac{10}{3}\mathbf{b}$ (Combine the $\mathbf{a}$ terms)

Answer: $\vec{OP} = \frac{2}{3}\mathbf{a} + \frac{10}{3}\mathbf{b}$

Worked Example 3 — Collinearity Proof

Question: The position vectors of points $A$, $B$, and $C$ relative to an origin $O$ are $\mathbf{a} = \begin{pmatrix} 1 \ 2 \end{pmatrix}$, $\mathbf{b} = \begin{pmatrix} 4 \ 8 \end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix} 7 \ 14 \end{pmatrix}$ respectively. Show that $A$, $B$, and $C$ are collinear.

1. Find $\vec{AB}$: $\vec{AB} = \mathbf{b} - \mathbf{a}$ (Vector between two points)
2. Substitute the position vectors: $\vec{AB} = \begin{pmatrix} 4 \ 8 \end{pmatrix} - \begin{pmatrix} 1 \ 2 \end{pmatrix}$ (Substitute $\mathbf{b}$ and $\mathbf{a}$)
3. Calculate the vector: $\vec{AB} = \begin{pmatrix} 3 \ 6 \end{pmatrix}$ (Subtract the components)
4. Find $\vec{BC}$: $\vec{BC} = \mathbf{c} - \mathbf{b}$ (Vector between two points)
5. Substitute the position vectors: $\vec{BC} = \begin{pmatrix} 7 \ 14 \end{pmatrix} - \begin{pmatrix} 4 \ 8 \end{pmatrix}$ (Substitute $\mathbf{c}$ and $\mathbf{b}$)
6. Calculate the vector: $\vec{BC} = \begin{pmatrix} 3 \ 6 \end{pmatrix}$ (Subtract the components)
7. Observe the relationship: $\vec{AB} = \vec{BC}$ (Both vectors are equal)
8. Conclude collinearity: Since $\vec{AB} = \vec{BC}$, the points $A$, $B$, and $C$ are collinear. They share a common point $B$ and have the same direction vector.

Answer: $A$, $B$, and $C$ are collinear.

D. Velocity and Resultant Vectors

Velocity is a vector. To find the position $\mathbf{r}$ of a particle after $t$ seconds: $$\mathbf{r} = \mathbf{r_0} + \mathbf{v}t$$ Where $\mathbf{r_0}$ is the initial position and $\mathbf{v}$ is the constant velocity.

Worked Example 4 (Particle Collision):

Question: Particle $A$ starts at $\begin{pmatrix} -2 \ 5 \end{pmatrix}$ with velocity $\begin{pmatrix} 3 \ -1 \end{pmatrix}$. Particle $B$ starts at $\begin{pmatrix} 10 \ -7 \end{pmatrix}$ with velocity $\begin{pmatrix} -1 \ 3 \end{pmatrix}$. Show they collide and find the time of collision.

1. Position of $A$ at time $t$: $\mathbf{r_A} = \mathbf{r_{0A}} + t\mathbf{v_A}$ (Apply the position formula)
2. Substitute the initial position and velocity of $A$: $\mathbf{r_A} = \begin{pmatrix} -2 \ 5 \end{pmatrix} + t\begin{pmatrix} 3 \ -1 \end{pmatrix}$ (Substitute $\mathbf{r_{0A}} = \begin{pmatrix} -2 \ 5 \end{pmatrix}$ and $\mathbf{v_A} = \begin{pmatrix} 3 \ -1 \end{pmatrix}$)
3. Simplify: $\mathbf{r_A} = \begin{pmatrix} -2 + 3t \ 5 - t \end{pmatrix}$ (Multiply the velocity vector by $t$ and add to the initial position)
4. Position of $B$ at time $t$: $\mathbf{r_B} = \mathbf{r_{0B}} + t\mathbf{v_B}$ (Apply the position formula)
5. Substitute the initial position and velocity of $B$: $\mathbf{r_B} = \begin{pmatrix} 10 \ -7 \end{pmatrix} + t\begin{pmatrix} -1 \ 3 \end{pmatrix}$ (Substitute $\mathbf{r_{0B}} = \begin{pmatrix} 10 \ -7 \end{pmatrix}$ and $\mathbf{v_B} = \begin{pmatrix} -1 \ 3 \end{pmatrix}$)
6. Simplify: $\mathbf{r_B} = \begin{pmatrix} 10 - t \ -7 + 3t \end{pmatrix}$ (Multiply the velocity vector by $t$ and add to the initial position)
7. Equate $x$-components to find the time of potential collision: $-2 + 3t = 10 - t$ (If they collide, their $x$-coordinates must be equal at the time of collision)
8. Solve for $t$: $4t = 12$ (Add $t$ and 2 to both sides)
9. Isolate $t$: $t = 3$ (Divide both sides by 4)
10. Check $y$-components with $t=3$: For $A$: $5 - (3) = 2$ (Substitute $t=3$ into the $y$-component of $\mathbf{r_A}$) For $B$: $-7 + 3(3) = 2$ (Substitute $t=3$ into the $y$-component of $\mathbf{r_B}$)
11. Since both components match at $t=3$, they collide at position $\begin{pmatrix} 7 \ 2 \end{pmatrix}$ at $t=3$ seconds.

Answer: The particles collide at $t=3$ seconds at the position $\begin{pmatrix} 7 \ 2 \end{pmatrix}$.

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Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations

$|\mathbf{a}| = \sqrt{x^2 + y^2}$ (Magnitude of a vector; use for speed calculation)

$\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}$ (Unit vector; magnitude is always 1)

$\vec{AB} = \mathbf{b} - \mathbf{a}$ (Vector between two points; essential for geometry)

$\mathbf{r} = \mathbf{r_0} + \mathbf{v}t$ (Position at time $t$; $\mathbf{r_0}$ is initial position)

$Speed = |\mathbf{v}|$ (Magnitude of velocity; always a non-negative scalar)

Note: These formulas are not provided on the formula sheet. You must memorize them.

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Common Mistakes to Avoid

• ❌ Wrong: Confusing the ratio $AC:AB = 1:3$ with $AC:CB = 1:3$ and incorrectly calculating the fraction of $\vec{AB}$.
• ✅ Right: Read carefully. If $AC:AB = 1:3$, then $\vec{AC} = \frac{1}{3}\vec{AB}$. If $AC:CB = 1:3$, then $\vec{AC} = \frac{1}{4}\vec{AB}$. Always relate back to the whole vector.
• ❌ Wrong: Calculating magnitude as $x^2 + y^2$ (forgetting the square root).
• ✅ Right: Magnitude is the hypotenuse of a triangle; always use $\sqrt{x^2 + y^2}$.
• ❌ Wrong: Using "Gradient" methods for vector proofs of parallelism.
• ✅ Right: Use scalar multiples. To show $AB$ is parallel to $CD$, show $\vec{AB} = k\vec{CD}$ for some scalar $k$.
• ❌ Wrong: Giving a velocity vector when asked for speed.
• ✅ Right: Velocity is $\begin{pmatrix} x \ y \end{pmatrix}$; Speed is $\sqrt{x^2+y^2}$. Speed is a scalar (magnitude only), so it cannot be negative.
• ❌ Wrong: Premature rounding in magnitude calculations, leading to inaccurate final answers.
• ✅ Right: Keep all intermediate values in exact form (surds, fractions) until the very last step, then round to the specified degree of accuracy.
• ❌ Wrong: Forgetting the $\pm$ when finding a unit vector parallel to a given line. There are two possible directions.
• ✅ Right: Remember that a line can be traversed in two directions. If the question doesn't specify a direction, provide both unit vectors: $\pm \frac{\mathbf{a}}{|\mathbf{a}|}$.

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Exam Tips

• Equating Scalars: In complex geometry questions where you have two expressions for the same vector (e.g., $\vec{OX} = \mu \mathbf{a} + \lambda \mathbf{b}$ and $\vec{OX} = 3\mathbf{a} + 4\mathbf{b}$), you must equate the components: $\mu = 3$ and $\lambda = 4$. This is a high-mark step.
• "Show That" Questions: Always state the general formula before substituting values. For example, write $\vec{AB} = \vec{OB} - \vec{OA}$ before doing the subtraction. This shows the examiner you understand the underlying principle.
• Non-Calculator (Paper 1): You will often get vectors with components like $\sqrt{3}$ or $1$. If asked for a unit vector, keep the denominator as a surd (e.g., $\frac{1}{\sqrt{5}}\begin{pmatrix} 1 \ 2 \end{pmatrix}$) rather than a decimal. Rationalize the denominator only if explicitly asked.
• Context: Velocity problems often involve ships or planes. Remember: "Resultant Velocity" = "Velocity in Still Water" + "Velocity of Current/Wind". Pay close attention to the directions given (e.g., bearings).
• Notation Check: In the exam, use $\underline{a}$ or $\vec{a}$ if you cannot write in bold. Examiners need to see you distinguish between a scalar $k$ and a vector $\mathbf{a}$.
• Diagrams: Always sketch a diagram, even if one is provided. This helps visualize the problem and identify the correct vector paths. Label all points and vectors clearly.
• Units: Remember to include units in your final answer, especially in velocity and position problems (e.g., $ms^{-1}$ for speed, $m$ for position).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

The position vectors of points $A$ and $B$ relative to an origin $O$ are given by $\overrightarrow{OA} = \begin{pmatrix} 5 \ -3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -2 \ 11 \end{pmatrix}$.

(a) Find $\overrightarrow{AB}$. [2]

(b) Find $|\overrightarrow{AB}|$, giving your answer in the form $k\sqrt{2}$, where $k$ is an integer. [3]

(c) Find the unit vector in the direction of $\overrightarrow{AB}$. [3]

Worked Solution:

(a)

1. Use the fact that $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ $\overrightarrow{AB} = \begin{pmatrix} -2 \ 11 \end{pmatrix} - \begin{pmatrix} 5 \ -3 \end{pmatrix}$ [Recognizing the correct vector subtraction]

2. Calculate the components. $\overrightarrow{AB} = \begin{pmatrix} -2-5 \ 11-(-3) \end{pmatrix} = \begin{pmatrix} -7 \ 14 \end{pmatrix}$

(b)

1. Find the magnitude of $\overrightarrow{AB}$ using Pythagoras' theorem: $|\overrightarrow{AB}| = \sqrt{(-7)^2 + (14)^2}$ [Correct application of Pythagoras]

2. Simplify the expression: $|\overrightarrow{AB}| = \sqrt{49 + 196} = \sqrt{245}$

3. Express in the form $k\sqrt{2}$: $|\overrightarrow{AB}| = \sqrt{49 \times 5} = 7\sqrt{5}$ [Correctly simplifying the surd]

(c)

1. Find the unit vector by dividing the vector by its magnitude: $\hat{u} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|} = \frac{1}{7\sqrt{5}} \begin{pmatrix} -7 \ 14 \end{pmatrix}$ [Recognizing to divide by magnitude]

2. Simplify each component: $\hat{u} = \begin{pmatrix} \frac{-7}{7\sqrt{5}} \ \frac{14}{7\sqrt{5}} \end{pmatrix} = \begin{pmatrix} \frac{-1}{\sqrt{5}} \ \frac{2}{\sqrt{5}} \end{pmatrix}$

3. Rationalize the denominator: $\hat{u} = \begin{pmatrix} \frac{-\sqrt{5}}{5} \ \frac{2\sqrt{5}}{5} \end{pmatrix}$

Final Answers: (a) $\overrightarrow{AB} = \boxed{\begin{pmatrix} -7 \ 14 \end{pmatrix}}$ (b) $|\overrightarrow{AB}| = \boxed{7\sqrt{5}}$ (c) $\hat{u} = \boxed{\begin{pmatrix} \frac{-\sqrt{5}}{5} \ \frac{2\sqrt{5}}{5} \end{pmatrix}}$

Common Pitfall: Remember that a unit vector must have a magnitude of 1. Always double-check your final answer to make sure you've fully simplified the surds and rationalized the denominator. Also, be careful with your signs when subtracting vectors.

How to earn full marks: For part (a), make sure you subtract the position vectors in the correct order (OB - OA). For part (b), show your working for simplifying the surd. For part (c), remember to rationalize the denominator in your final answer.

```markdown
#### Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [7 marks]

**Question:**

The vectors $\mathbf{a}$ and $\mathbf{b}$ are given by $\mathbf{a} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -4 \\ 1 \end{pmatrix}$.

(a) Find $3\mathbf{a} - \mathbf{b}$. [2]

(b) Find the scalar $k$ such that $3\mathbf{a} - \mathbf{b}$ is parallel to the vector $\begin{pmatrix} 1 \\ k \end{pmatrix}$. [5]

**Worked Solution:**

**(a)**
1. Calculate $3\mathbf{a}$:
   $3\mathbf{a} = 3\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 6 \\ -9 \end{pmatrix}$
   *[Correct scalar multiplication]*

2. Calculate $3\mathbf{a} - \mathbf{b}$:
   $3\mathbf{a} - \mathbf{b} = \begin{pmatrix} 6 \\ -9 \end{pmatrix} - \begin{pmatrix} -4 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 - (-4) \\ -9 - 1 \end{pmatrix} = \begin{pmatrix} 10 \\ -10 \end{pmatrix}$
   *[Correct vector subtraction]*

   

**(b)**
1. State the condition for parallel vectors:
   If $3\mathbf{a} - \mathbf{b}$ is parallel to $\begin{pmatrix} 1 \\ k \end{pmatrix}$, then $\begin{pmatrix} 10 \\ -10 \end{pmatrix} = \lambda \begin{pmatrix} 1 \\ k \end{pmatrix}$ for some scalar $\lambda$.
   *[Understanding parallel vectors are scalar multiples]*

2. Equate the x-components:
   $10 = \lambda(1)$, so $\lambda = 10$
   *[Finding the value of lambda]*

3. Equate the y-components:
   $-10 = \lambda k = 10k$
   *[Setting up equation to find k]*

4. Solve for $k$:
   $k = \frac{-10}{10} = -1$
   *[Correctly solving for k]*

   

Final Answers:
**(a)** $3\mathbf{a} - \mathbf{b} = \boxed{\begin{pmatrix} 10 \\ -10 \end{pmatrix}}$
**(b)** $k = \boxed{-1}$

**Common Pitfall:** When dealing with parallel vectors, remember that one vector is a scalar multiple of the other. Don't try to equate the vectors directly without introducing a scalar. Also, pay close attention to the order of operations when performing scalar multiplication and vector subtraction.

**How to earn full marks:** For part (a), show each step of the scalar multiplication and vector subtraction. For part (b), clearly state that parallel vectors are scalar multiples of each other and show how you found lambda.

#### Exam-Style Question 3 — Paper 2 (Calculator Allowed) [7 marks]

**Question:**

A particle moves such that its position vector at time $t$ seconds is given by $\mathbf{r} = \begin{pmatrix} 1 \\ -2 \end{pmatrix} + t\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, where the components are in meters.

(a) Find the position vector of the particle when $t=4$. [2]

(b) Find the speed of the particle. [3]

(c) Find the time at which the particle is 15 meters from the origin. [2]

**Worked Solution:**

**(a)**
1. Substitute $t=4$ into the expression for $\mathbf{r}$:
   $\mathbf{r} = \begin{pmatrix} 1 \\ -2 \end{pmatrix} + 4\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \end{pmatrix} + \begin{pmatrix} 16 \\ 12 \end{pmatrix}$
   *[Correct substitution]*

2. Calculate the position vector:
   $\mathbf{r} = \begin{pmatrix} 1+16 \\ -2+12 \end{pmatrix} = \begin{pmatrix} 17 \\ 10 \end{pmatrix}$
   *[Correct vector addition]*

   

**(b)**
1. Identify the velocity vector: The velocity vector is the coefficient of $t$, so $\mathbf{v} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$.
   *[Recognizing velocity vector]*

2. Find the speed, which is the magnitude of the velocity vector:
   $|\mathbf{v}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$
   *[Correct application of Pythagoras]*

3. State the units: The speed is 5 m/s.

   

**(c)**
1. Write the general position vector as $\mathbf{r} = \begin{pmatrix} 1+4t \\ -2+3t \end{pmatrix}$
   *[Writing position vector in component form]*

2. Set up the equation $|\mathbf{r}| = 15$:
   $\sqrt{(1+4t)^2 + (-2+3t)^2} = 15$
   *[Correct setup of the magnitude equation]*

3. Square both sides:
   $(1+4t)^2 + (-2+3t)^2 = 225$
   $1 + 8t + 16t^2 + 4 - 12t + 9t^2 = 225$
   $25t^2 - 4t + 5 = 225$
   $25t^2 - 4t - 220 = 0$
   *[Expanding and rearranging to form a quadratic]*

4. Use quadratic formula to solve for $t$:
   $t = \frac{4 \pm \sqrt{(-4)^2 - 4(25)(-220)}}{2(25)} = \frac{4 \pm \sqrt{16 + 22000}}{50} = \frac{4 \pm \sqrt{22016}}{50} = \frac{4 \pm 148.378}{50}$
   *[Correct use of quadratic formula]*

5. The two possible values for $t$ are $t = \frac{4 + 148.378}{50} \approx 3.048$ and $t = \frac{4 - 148.378}{50} \approx -2.888$.  Since time cannot be negative, $t \approx 3.05$
   *[Selecting the positive root]*

   

Final Answers:
**(a)** $\mathbf{r} = \boxed{\begin{pmatrix} 17 \\ 10 \end{pmatrix}}$
**(b)** Speed = $\boxed{5 \text{ m/s}}$
**(c)** $t = \boxed{3.05 \text{ seconds (to 3 s.f.)}}$

**Common Pitfall:** Remember that speed is the magnitude of the velocity vector, not the position vector. Also, when solving for time, make sure to discard any negative solutions, as time cannot be negative in this context. Be careful when expanding the brackets and using the quadratic formula, as small errors can lead to incorrect answers.

**How to earn full marks:** For part (a), show the scalar multiplication and vector addition. For part (b), remember to include the units (m/s) in your final answer. For part (c), show your working when using the quadratic formula and state the answer to 3 significant figures.

#### Exam-Style Question 4 — Paper 2 (Calculator Allowed) [8 marks]

**Question:**

Points $A$, $B$, and $C$ have position vectors $\overrightarrow{OA} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix} 10 \\ -2 \end{pmatrix}$, and $\overrightarrow{OC} = \begin{pmatrix} 1 \\ 7 \end{pmatrix}$ respectively, relative to an origin $O$.

(a) Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$. [2]

(b) Find the angle $BAC$. [4]

(c) Find the area of triangle $ABC$. [2]

**Worked Solution:**

**(a)**
1. Find $\overrightarrow{AB}$:
   $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 10 \\ -2 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}$
   *[Correct vector subtraction]*

2. Find $\overrightarrow{AC}$:
   $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 1 \\ 7 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \end{pmatrix}$
   *[Correct vector subtraction]*

   

**(b)**
1. Use the cosine rule for vectors: $\overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}| |\overrightarrow{AC}| \cos{\angle BAC}$
   *[Recognizing correct formula]*

2. Calculate the dot product $\overrightarrow{AB} \cdot \overrightarrow{AC}$:
   $\overrightarrow{AB} \cdot \overrightarrow{AC} = (6)(-3) + (-3)(6) = -18 - 18 = -36$
   *[Correct dot product calculation]*

3. Calculate the magnitudes $|\overrightarrow{AB}|$ and $|\overrightarrow{AC}|$:
   $|\overrightarrow{AB}| = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$
   $|\overrightarrow{AC}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$
   *[Correct magnitude calculations]*

4. Substitute into the cosine rule formula and solve for $\cos{\angle BAC}$:
   $-36 = (3\sqrt{5})(3\sqrt{5}) \cos{\angle BAC}$
   $-36 = 45 \cos{\angle BAC}$
   $\cos{\angle BAC} = \frac{-36}{45} = -\frac{4}{5}$
   *[Correct substitution and simplification]*

5. Find $\angle BAC$:
   $\angle BAC = \cos^{-1}\left(-\frac{4}{5}\right) \approx 2.214 \text{ radians or } 126.87^{\circ}$

   

**(c)**
1. Use the formula for the area of a triangle: Area $= \frac{1}{2} |\overrightarrow{AB}| |\overrightarrow{AC}| \sin{\angle BAC}$
    *[Recognizing the area formula]*

2. Calculate $\sin{\angle BAC}$:
    Since $\cos{\angle BAC} = -\frac{4}{5}$, we can use the identity $\sin^2{\theta} + \cos^2{\theta} = 1$.
    $\sin^2{\angle BAC} = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}$
    $\sin{\angle BAC} = \sqrt{\frac{9}{25}} = \frac{3}{5}$ (Since the angle is obtuse, sine is positive)
    *[Correctly finding sin(angle)]*

3. Substitute to find the area:
    Area $= \frac{1}{2} (3\sqrt{5})(3\sqrt{5}) \left(\frac{3}{5}\right) = \frac{1}{2} (45) \left(\frac{3}{5}\right) = \frac{1}{2} (9)(3) = \frac{27}{2} = 13.5$
    *[Correctly calculating the area]*

   

Final Answers:
**(a)** $\overrightarrow{AB} = \boxed{\begin{pmatrix} 6 \\ -3 \end{pmatrix}}$, $\overrightarrow{AC} = \boxed{\begin{pmatrix} -3 \\ 6 \end{pmatrix}}$
**(b)** $\angle BAC = \boxed{126.9^{\circ} \text{ (to 1 d.p.) or } 2.21 \text{ radians (to 3 s.f.)}}$
**(c)** Area $= \boxed{13.5 \text{ units}^2}$

**Common Pitfall:** When finding the angle between two vectors, make sure you are using the correct vectors (in this case, $\overrightarrow{AB}$ and $\overrightarrow{AC}$). A common mistake is to use $\overrightarrow{BA}$ instead of $\overrightarrow{AB}$, which will result in an incorrect angle. Also, remember to use the correct formula for the area of a triangle when given two sides and the included angle.

**How to earn full marks:** For part (a), show the vector subtraction clearly. For part (b), state the cosine rule formula before substituting and give the answer in either degrees or radians. For part (c), show how you calculated sin(angle) and remember to include the units in your final answer.