| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with line |
| Difficulty | Standard +0.3 This is a standard three-dimensional vectors question requiring equating parametric equations to find intersection conditions, using dot product for perpendicularity, and solving simultaneous equations. While it involves multiple steps across three parts, each technique is routine for Further Maths students and follows predictable methods without requiring novel insight or complex geometric reasoning. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Express general point of \(l\) or \(m\) in component form: \((-1+2\lambda,3-\lambda,4-\lambda)\) or \((5+a\mu,4+b\mu,3+\mu)\) | B1 | |
| Equate components and eliminate either \(\lambda\) or \(\mu\) | M1 | e.g. \(\mu=\frac{2}{1-b}\), \(\lambda=\frac{1-b}{b}\), \(\mu=\frac{-4}{2+a}\), \(\lambda=\frac{a+6}{a+2}\) |
| Eliminate the other parameter or obtain a second expression in the first | M1 | \(\lambda\) and \(\mu\) not required to be subject |
| Show intermediate steps to obtain \(2b-a=4\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Express general point of \(l\) or \(m\) in component form | B1 | |
| Express \(a\) or \(b\) in terms of \(\lambda\) and \(\mu\) | M1 | \(a=\dfrac{2\lambda-6}{\mu}\), \(b=\dfrac{-1-\lambda}{\mu}\) |
| Use \(\lambda=1-\mu\) | M1 | |
| Obtain \(2b-a=4\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using correct process, equate scalar product of direction vectors to zero | \*M1 | \((2\mathbf{i}-\mathbf{j}-\mathbf{k})\cdot(a\mathbf{i}+b\mathbf{j}+\mathbf{k})=0\) SOI |
| Obtain \(2a-b-1=0\) | A1 | OE e.g. \(2(2b-4)-b-1=0\) |
| Solve simultaneous equations for \(a\) or for \(b\) | DM1 | |
| Obtain \(a=2\), \(b=3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute found values in component equations and solve for \(\lambda\) or \(\mu\) | M1 | |
| Obtain answer \(3\mathbf{i}+\mathbf{j}+2\mathbf{k}\) from either \(\lambda=2\) or \(\mu=-1\) | A1 | Accept as coordinates or equivalent |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Express general point of $l$ or $m$ in component form: $(-1+2\lambda,3-\lambda,4-\lambda)$ or $(5+a\mu,4+b\mu,3+\mu)$ | B1 | |
| Equate components and eliminate either $\lambda$ or $\mu$ | M1 | e.g. $\mu=\frac{2}{1-b}$, $\lambda=\frac{1-b}{b}$, $\mu=\frac{-4}{2+a}$, $\lambda=\frac{a+6}{a+2}$ |
| Eliminate the other parameter or obtain a second expression in the first | M1 | $\lambda$ and $\mu$ not required to be subject |
| Show intermediate steps to obtain $2b-a=4$ | A1 | AG |
**Alternative method for 9(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Express general point of $l$ or $m$ in component form | B1 | |
| Express $a$ or $b$ in terms of $\lambda$ and $\mu$ | M1 | $a=\dfrac{2\lambda-6}{\mu}$, $b=\dfrac{-1-\lambda}{\mu}$ |
| Use $\lambda=1-\mu$ | M1 | |
| Obtain $2b-a=4$ | A1 | AG |
**Total: 4 marks**
---
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Using correct process, equate scalar product of direction vectors to zero | \*M1 | $(2\mathbf{i}-\mathbf{j}-\mathbf{k})\cdot(a\mathbf{i}+b\mathbf{j}+\mathbf{k})=0$ SOI |
| Obtain $2a-b-1=0$ | A1 | OE e.g. $2(2b-4)-b-1=0$ |
| Solve simultaneous equations for $a$ or for $b$ | DM1 | |
| Obtain $a=2$, $b=3$ | A1 | |
**Total: 4 marks**
---
## Question 9(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute found values in component equations and solve for $\lambda$ or $\mu$ | M1 | |
| Obtain answer $3\mathbf{i}+\mathbf{j}+2\mathbf{k}$ from either $\lambda=2$ or $\mu=-1$ | A1 | Accept as coordinates or equivalent |
**Total: 2 marks**
---9 The lines $l$ and $m$ have vector equations
$$\mathbf { r } = - \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k } + \lambda ( 2 \mathbf { i } - \mathbf { j } - \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = 5 \mathbf { i } + 4 \mathbf { j } + 3 \mathbf { k } + \mu ( a \mathbf { i } + b \mathbf { j } + \mathbf { k } )$$
respectively, where $a$ and $b$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Given that $l$ and $m$ intersect, show that $2 b - a = 4$.
\item Given also that $l$ and $m$ are perpendicular, find the values of $a$ and $b$.
\item When $a$ and $b$ have these values, find the position vector of the point of intersection of $l$ and $m$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q9 [10]}}