| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Reflection in plane |
| Difficulty | Challenging +1.2 This is a structured multi-part vectors question requiring standard techniques: dot product for perpendicularity, solving simultaneous equations for intersection, parameter substitution to verify a point on a line, and reflection geometry. While part (d) requires careful geometric reasoning about reflection in a line (finding perpendicular from A to l₂, then reflecting), each step follows established methods without requiring novel insight. The question is moderately harder than average due to the reflection component and multi-step nature, but remains within standard Further Maths Pure territory. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Carry out correct process for evaluating the scalar product of direction vectors, equate to zero and obtain \(a = 4\) | B1 | E.g. \(2(3) + (-1)(-2) + a(-2) = 0\) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Express general point of at least one line correctly in component form, i.e. \((1+2\lambda, -2-\lambda, 3+4\lambda)\) or \((-1+3\mu, -1-2\mu, -1-2\mu)\) | B1 | Third component could be implied by a correct final answer |
| Equate at least two pairs of corresponding components and solve for \(\lambda\) or \(\mu\) | M1 | |
| Obtain \(\lambda = -1\) or \(\mu = 0\) | A1 | |
| Obtain position vector of point of intersection \(-\mathbf{i} - \mathbf{j} - \mathbf{k}\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate one component of \(l_1\) to matching component of \(A\) and solve to find \(\lambda\) | M1 | |
| Use \(\lambda = -3\) in equation of \(l_1\) and show this gives position vector of \(A\) | A1 | AG. Or show \(\lambda = -3\) for all three components equated |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Method to find position vector of \(B\) | M1 | E.g. \(\pm 2 \times their\ (-\mathbf{i}-\mathbf{j}-\mathbf{k}) \pm (-5\mathbf{i}+\mathbf{j}-9\mathbf{k})\) |
| Obtain position vector of \(B\) is \(3\mathbf{i} - 3\mathbf{j} + 7\mathbf{k}\) | A1 | |
| 2 |
## Question 9:
**Part 9(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Carry out correct process for evaluating the scalar product of direction vectors, equate to zero and obtain $a = 4$ | B1 | E.g. $2(3) + (-1)(-2) + a(-2) = 0$ |
| | **1** | |
**Part 9(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Express general point of at least one line correctly in component form, i.e. $(1+2\lambda, -2-\lambda, 3+4\lambda)$ or $(-1+3\mu, -1-2\mu, -1-2\mu)$ | B1 | Third component could be implied by a correct final answer |
| Equate at least two pairs of corresponding components and solve for $\lambda$ or $\mu$ | M1 | |
| Obtain $\lambda = -1$ or $\mu = 0$ | A1 | |
| Obtain position vector of point of intersection $-\mathbf{i} - \mathbf{j} - \mathbf{k}$ | A1 | |
| | **4** | |
**Part 9(c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate one component of $l_1$ to matching component of $A$ and solve to find $\lambda$ | M1 | |
| Use $\lambda = -3$ in equation of $l_1$ and show this gives position vector of $A$ | A1 | AG. Or show $\lambda = -3$ for all three components equated |
| | **2** | |
**Part 9(d):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Method to find position vector of $B$ | M1 | E.g. $\pm 2 \times their\ (-\mathbf{i}-\mathbf{j}-\mathbf{k}) \pm (-5\mathbf{i}+\mathbf{j}-9\mathbf{k})$ |
| Obtain position vector of $B$ is $3\mathbf{i} - 3\mathbf{j} + 7\mathbf{k}$ | A1 | |
| | **2** | |
---9 The equations of two straight lines $l _ { 1 }$ and $l _ { 2 }$ are
$$l _ { 1 } : \quad \mathbf { r } = \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } + \lambda ( 2 \mathbf { i } - \mathbf { j } + a \mathbf { k } ) \quad \text { and } \quad l _ { 2 } : \quad \mathbf { r } = - \mathbf { i } - \mathbf { j } - \mathbf { k } + \mu ( 3 \mathbf { i } - 2 \mathbf { j } - 2 \mathbf { k } ) ,$$
where $a$ is a constant.\\
The lines $l _ { 1 }$ and $l _ { 2 }$ are perpendicular.
\begin{enumerate}[label=(\alph*)]
\item Show that $a = 4$.\\
The lines $l _ { 1 }$ and $l _ { 2 }$ also intersect.
\item Find the position vector of the point of intersection.\\
The point $A$ has position vector $- 5 \mathbf { i } + \mathbf { j } - 9 \mathbf { k }$.
\item Show that $A$ lies on $l _ { 1 }$.\\
The point $B$ is the image of $A$ after a reflection in the line $l _ { 2 }$.
\item Find the position vector of $B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q9 [9]}}