| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Triangle and parallelogram areas |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: identifying intersection point from equations, finding angle between lines using dot product, and calculating triangle area using cross product. All methods are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2, 1, -3)\) | B1 | Accept \(x=2\), \(y=1\), \(z=-3\). Do not accept \(2\mathbf{i} + \mathbf{j} - 3\mathbf{k}\) or column vector form. |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use the correct method to find the scalar product of the direction vectors | M1 | \(\pm(-1\times2 + 2\times5) = \pm8\). Allow error of \(0\times-1=-1\). |
| Divide the scalar product by the product of the moduli to obtain \(\pm\cos\theta\) using consistent vectors throughout | M1 | \(\dfrac{\text{their } 8}{\sqrt{\text{their } 5}\sqrt{\text{their } 30}}\) |
| Obtain \(\cos\theta = \dfrac{8}{5\sqrt{6}}\) | A1 | OE e.g. \(\dfrac{8}{\sqrt{150}}\) or \(\dfrac{4\sqrt{6}}{15}\). If no \(\dfrac{8}{5\sqrt{6}}\) seen, just 49.2, then A0. Decimal only seen, A0. ISW. |
| Alternative Method: Use of cosine rule: sides of \(\sqrt{5}\), \(\sqrt{30}\) and \(\sqrt{19}\) found | B1 | Could use other points. |
| e.g. \(\cos\theta = \dfrac{5 + 30 - 19}{2\sqrt{5}\sqrt{30}}\) | M1 | |
| Obtain \(\cos\theta = \dfrac{8}{5\sqrt{6}}\) | A1 | OE. Decimal only seen, A0. ISW. |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Any two of \( | PA | = 2\sqrt{5}\), \( |
| Area \(= \dfrac{1}{2} \times 2\sqrt{5} \times \sqrt{30} \times \sqrt{1 - \dfrac{64}{150}}\) | M1 | Correct method for exact area. Note: \(\sin APB = \dfrac{\sqrt{129}}{15}\), \(\sin ABP = \sqrt{\dfrac{86}{615}}\), \(\cos ABP = \dfrac{46}{\sqrt{2460}}\), Perp \(A\) to \(BP = \dfrac{\sqrt{2580}}{15}\), Perp \(B\) to \(AP = \dfrac{\sqrt{430}}{5}\) |
| \(= \sqrt{86}\) | A1 | Or simplified exact equivalent. ISW. |
| Alternative Method: \(\vec{PA} \times \vec{PB} = -4\mathbf{i} - 18\mathbf{j} - 2\mathbf{k}\) | B1 | \(\vec{PA} = -2\mathbf{i} + 4\mathbf{k}\), \(\vec{PB} = -2\mathbf{i} + \mathbf{j} - 5\mathbf{k}\) |
| Area \(= \dfrac{1}{2} | \vec{PA} \times \vec{PB} | = \dfrac{1}{2}\sqrt{16 + 324 + 4}\) |
| \(= \sqrt{86}\) | A1 | Or simplified exact equivalent. ISW. |
| Total | 3 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2, 1, -3)$ | B1 | Accept $x=2$, $y=1$, $z=-3$. Do not accept $2\mathbf{i} + \mathbf{j} - 3\mathbf{k}$ or column vector form. |
| **Total** | **1** | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use the correct method to find the scalar product of the direction vectors | M1 | $\pm(-1\times2 + 2\times5) = \pm8$. Allow error of $0\times-1=-1$. |
| Divide the scalar product by the product of the moduli to obtain $\pm\cos\theta$ using consistent vectors throughout | M1 | $\dfrac{\text{their } 8}{\sqrt{\text{their } 5}\sqrt{\text{their } 30}}$ |
| Obtain $\cos\theta = \dfrac{8}{5\sqrt{6}}$ | A1 | OE e.g. $\dfrac{8}{\sqrt{150}}$ or $\dfrac{4\sqrt{6}}{15}$. If no $\dfrac{8}{5\sqrt{6}}$ seen, just 49.2, then A0. Decimal only seen, A0. ISW. |
| **Alternative Method:** Use of cosine rule: sides of $\sqrt{5}$, $\sqrt{30}$ and $\sqrt{19}$ found | B1 | Could use other points. |
| e.g. $\cos\theta = \dfrac{5 + 30 - 19}{2\sqrt{5}\sqrt{30}}$ | M1 | |
| Obtain $\cos\theta = \dfrac{8}{5\sqrt{6}}$ | A1 | OE. Decimal only seen, A0. ISW. |
| **Total** | **3** | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Any two of $|PA| = 2\sqrt{5}$, $|PB| = \sqrt{30}$, or $|AB| = \sqrt{82}$ seen | B1 | May be seen by stating or implying $\lambda = 2$ and $\mu = -1$. |
| Area $= \dfrac{1}{2} \times 2\sqrt{5} \times \sqrt{30} \times \sqrt{1 - \dfrac{64}{150}}$ | M1 | Correct method for exact area. Note: $\sin APB = \dfrac{\sqrt{129}}{15}$, $\sin ABP = \sqrt{\dfrac{86}{615}}$, $\cos ABP = \dfrac{46}{\sqrt{2460}}$, Perp $A$ to $BP = \dfrac{\sqrt{2580}}{15}$, Perp $B$ to $AP = \dfrac{\sqrt{430}}{5}$ |
| $= \sqrt{86}$ | A1 | Or simplified exact equivalent. ISW. |
| **Alternative Method:** $\vec{PA} \times \vec{PB} = -4\mathbf{i} - 18\mathbf{j} - 2\mathbf{k}$ | B1 | $\vec{PA} = -2\mathbf{i} + 4\mathbf{k}$, $\vec{PB} = -2\mathbf{i} + \mathbf{j} - 5\mathbf{k}$ |
| Area $= \dfrac{1}{2}|\vec{PA} \times \vec{PB}| = \dfrac{1}{2}\sqrt{16 + 324 + 4}$ | M1 | Correct method for exact area. |
| $= \sqrt{86}$ | A1 | Or simplified exact equivalent. ISW. |
| **Total** | **3** | |
---6 The lines $l$ and $m$ have vector equations
$$l : \quad \mathbf { r } = 2 \mathbf { i } + \mathbf { j } - 3 \mathbf { k } + \lambda ( - \mathbf { i } + 2 \mathbf { k } ) \quad \text { and } \quad m : \quad \mathbf { r } = 2 \mathbf { i } + \mathbf { j } - 3 \mathbf { k } + \mu ( 2 \mathbf { i } - \mathbf { j } + 5 \mathbf { k } ) .$$
Lines $l$ and $m$ intersect at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of $P$.
\item Find the exact value of the cosine of the acute angle between $l$ and $m$.
\item The point $A$ on line $I$ has coordinates ( $0,1,1$ ). The point $B$ on line $m$ has coordinates ( $0,2 , - 8$ ).
Find the exact area of triangle $A P B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q6 [7]}}