| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.3 This is a standard vectors question requiring perpendicularity condition (dot product = 0), point-on-line substitution, and ratio division along a line. Part (a) involves routine algebraic manipulation with three equations in three unknowns. Part (b) requires finding the foot of perpendicular and applying section formula—all standard techniques covered in P3 with no novel insight required. Slightly easier than average due to straightforward application of methods. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Perform scalar product of direction vectors and set result equal to zero | M1 | \(2c + 6 + 4 = 0\) |
| Use \(P\) to find the value of \(\lambda\) | M1 | \(3 - 2\lambda = 7 \Rightarrow \lambda = -2\) \([a + \lambda c = 4, b + 4\lambda = -2]\). Equation for line \(l\) may contain \(-\lambda\) instead of \(+\lambda\) leading to \(\lambda = 2\), all marks available. |
| Obtain \(c = -5\) or \(b = 6\) | A1 | |
| \(a = -6\), \(b = 6\) and \(c = -5\) all correct | A1 | SC1: Use \(P\) to find value of \(\lambda\) M1. Substitute \(\lambda = -2\) into point \(P\), so \(a - 2c = 4\), and put \(\mu = -1\) and \(\lambda = -1\) into \(l\) so \(a - c = -1\), then solve to obtain \(a = -6\), \(b = 6\) and \(c = -5\). All 3 values correct A1. Max 2/4. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Find \(\overrightarrow{PQ}\) (or \(\overrightarrow{QP}\)) for a general point \(Q\) on \(m\) \(= \pm((1+2\mu, 2-3\mu, 3+\mu) - (a+\lambda c, 3-2\lambda, b+4\lambda))\) | B1 | \(\overrightarrow{PQ}\) or \(\overrightarrow{QP} = \pm\begin{pmatrix}-3+2\mu\\-5-3\mu\\5+\mu\end{pmatrix}\). Could be their \(a, b, c\) and \(\lambda\) values provided M1 M1 gained in (a). Allow expression in answer column. |
| Equate scalar product of \(\overrightarrow{PQ}\) (or \(\overrightarrow{QP}\)) and a direction vector for \(m\) to zero and obtain an equation in \(\mu\) | M1* | \((2(-3+2\mu) - 3(-5-3\mu) + (5+\mu)) = 0\). Allow \(\overrightarrow{PQ} = \overrightarrow{OQ} + \overrightarrow{OP}\) sign problem. |
| Solve and obtain \(\mu = -1\) | A1 | \(PQ^2 = (-3+2\mu)^2 + (-5-3\mu)^2 + (5+\mu)^2\). \([= 14(\mu+1)^2 + 45]\). Min when \(\mu = -1\) or by differentiation. |
| Obtain \(\overrightarrow{OQ} = -\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\) or \(\overrightarrow{PQ} = -5\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\). Must be labelled correctly | A1 | The working may be in (a) provided at least this result is used in (b). |
| Carry out a method to find the position vector of \(R\) | DM1 | e.g. Use \(\overrightarrow{OR} = \overrightarrow{OP} + \frac{5}{2}\overrightarrow{PQ}\) or \(\overrightarrow{OR} = \overrightarrow{OQ} + \frac{3}{2}\overrightarrow{PQ}\) or \(\overrightarrow{OR} = \frac{5}{2}\overrightarrow{OQ} - \frac{3}{2}\overrightarrow{OP}\) or \(2\overrightarrow{QR} = 2(\overrightarrow{OR} - \overrightarrow{OQ}) = 3\overrightarrow{PQ}\). \(\overrightarrow{PQ}\) used in all these approaches may be incorrect, must be in correct direction, i.e. not using \(\overrightarrow{QP}\) for \(\overrightarrow{PQ}\). Alternative for DM1: \(\overrightarrow{OR} = (4,7,-2) + t(-5,-2,4)\), \(\overrightarrow{QR} = \overrightarrow{OR} - \overrightarrow{OQ}\). Solve \(\lvert QR\rvert^2 = \frac{9}{4}\lvert PQ\rvert^2\) or \(\lvert QR\rvert = \frac{3}{2}\lvert PQ\rvert\), \(t = 2.5\) |
| Obtain \(-\frac{17}{2}\mathbf{i} + 2\mathbf{j} + 8\mathbf{k}\) from correct working | A1 | Accept coordinates. Don't accept \(-\frac{17}{2}\mathbf{i} + \frac{4}{2}\mathbf{j} + \frac{16}{2}\mathbf{k}\). SC2: Equate lines, attempt to find \(\mu = -1\) or \(\lambda = -1\) M1*. \(\overrightarrow{OQ} = -\mathbf{i}+5\mathbf{j}+2\mathbf{k}\) A1. Attempt to find \(\overrightarrow{OQ}\) using other parameter value DM1. \(\overrightarrow{OQ} = -\mathbf{i}+5\mathbf{j}+2\mathbf{k}\) therefore intersect A1. Then use main scheme for final DM1 A1. First DM1 A1 available if they show 3 coordinates consistent for the 2 parameter values instead of attempting to find \(\overrightarrow{OQ}\) using the other parameter value and then showing intersection. |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Perform scalar product of direction vectors and set result equal to zero | **M1** | $2c + 6 + 4 = 0$ |
| Use $P$ to find the value of $\lambda$ | **M1** | $3 - 2\lambda = 7 \Rightarrow \lambda = -2$ $[a + \lambda c = 4, b + 4\lambda = -2]$. Equation for line $l$ may contain $-\lambda$ instead of $+\lambda$ leading to $\lambda = 2$, all marks available. |
| Obtain $c = -5$ or $b = 6$ | **A1** | |
| $a = -6$, $b = 6$ and $c = -5$ all correct | **A1** | **SC1**: Use $P$ to find value of $\lambda$ M1. Substitute $\lambda = -2$ into point $P$, so $a - 2c = 4$, and put $\mu = -1$ and $\lambda = -1$ into $l$ so $a - c = -1$, then solve to obtain $a = -6$, $b = 6$ and $c = -5$. All 3 values correct A1. Max 2/4. |
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## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Find $\overrightarrow{PQ}$ (or $\overrightarrow{QP}$) for a general point $Q$ on $m$ $= \pm((1+2\mu, 2-3\mu, 3+\mu) - (a+\lambda c, 3-2\lambda, b+4\lambda))$ | **B1** | $\overrightarrow{PQ}$ or $\overrightarrow{QP} = \pm\begin{pmatrix}-3+2\mu\\-5-3\mu\\5+\mu\end{pmatrix}$. Could be their $a, b, c$ and $\lambda$ values provided M1 M1 gained in (a). Allow expression in answer column. |
| Equate scalar product of $\overrightarrow{PQ}$ (or $\overrightarrow{QP}$) and a direction vector for $m$ to zero and obtain an equation in $\mu$ | **M1*** | $(2(-3+2\mu) - 3(-5-3\mu) + (5+\mu)) = 0$. Allow $\overrightarrow{PQ} = \overrightarrow{OQ} + \overrightarrow{OP}$ sign problem. |
| Solve and obtain $\mu = -1$ | **A1** | $PQ^2 = (-3+2\mu)^2 + (-5-3\mu)^2 + (5+\mu)^2$. $[= 14(\mu+1)^2 + 45]$. Min when $\mu = -1$ or by differentiation. |
| Obtain $\overrightarrow{OQ} = -\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}$ or $\overrightarrow{PQ} = -5\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}$. Must be labelled correctly | **A1** | The working may be in **(a)** provided at least this result is used in **(b)**. |
| Carry out a method to find the position vector of $R$ | **DM1** | e.g. Use $\overrightarrow{OR} = \overrightarrow{OP} + \frac{5}{2}\overrightarrow{PQ}$ or $\overrightarrow{OR} = \overrightarrow{OQ} + \frac{3}{2}\overrightarrow{PQ}$ or $\overrightarrow{OR} = \frac{5}{2}\overrightarrow{OQ} - \frac{3}{2}\overrightarrow{OP}$ or $2\overrightarrow{QR} = 2(\overrightarrow{OR} - \overrightarrow{OQ}) = 3\overrightarrow{PQ}$. $\overrightarrow{PQ}$ used in all these approaches may be incorrect, must be in correct direction, i.e. not using $\overrightarrow{QP}$ for $\overrightarrow{PQ}$. **Alternative for DM1**: $\overrightarrow{OR} = (4,7,-2) + t(-5,-2,4)$, $\overrightarrow{QR} = \overrightarrow{OR} - \overrightarrow{OQ}$. Solve $\lvert QR\rvert^2 = \frac{9}{4}\lvert PQ\rvert^2$ or $\lvert QR\rvert = \frac{3}{2}\lvert PQ\rvert$, $t = 2.5$ |
| Obtain $-\frac{17}{2}\mathbf{i} + 2\mathbf{j} + 8\mathbf{k}$ from correct working | **A1** | Accept coordinates. Don't accept $-\frac{17}{2}\mathbf{i} + \frac{4}{2}\mathbf{j} + \frac{16}{2}\mathbf{k}$. **SC2**: Equate lines, attempt to find $\mu = -1$ or $\lambda = -1$ M1*. $\overrightarrow{OQ} = -\mathbf{i}+5\mathbf{j}+2\mathbf{k}$ A1. Attempt to find $\overrightarrow{OQ}$ using other parameter value DM1. $\overrightarrow{OQ} = -\mathbf{i}+5\mathbf{j}+2\mathbf{k}$ therefore intersect A1. Then use main scheme for final DM1 A1. First **DM1 A1** available if they show 3 coordinates consistent for the 2 parameter values instead of attempting to find $\overrightarrow{OQ}$ using the other parameter value and then showing intersection. |
---9 The lines $l$ and $m$ have equations
$$\begin{aligned}
l : & \mathbf { r } = a \mathbf { i } + 3 \mathbf { j } + b \mathbf { k } + \lambda ( c \mathbf { i } - 2 \mathbf { j } + 4 \mathbf { k } ) \\
m : & \mathbf { r } = \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } + \mu ( 2 \mathbf { i } - 3 \mathbf { j } + \mathbf { k } )
\end{aligned}$$
Relative to the origin $O$, the position vector of the point $P$ is $4 \mathbf { i } + 7 \mathbf { j } - 2 \mathbf { k }$.
\begin{enumerate}[label=(\alph*)]
\item Given that $l$ is perpendicular to $m$ and that $P$ lies on $l$, find the values of the constants $a , b$ and $c$. [4]
\item The perpendicular from $P$ meets line $m$ at $Q$. The point $R$ lies on $P Q$ extended, with $P Q : Q R = 2 : 3$.
Find the position vector of $R$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q9 [10]}}