| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.8 This is a multi-part vectors question requiring vector manipulation, angle bisector properties, and ratio verification. Part (i) is routine, but part (ii) requires setting up and solving a non-trivial equation involving dot products and magnitudes, while part (iii) tests understanding of the angle bisector theorem. The conceptual demand and algebraic complexity place it above average difficulty. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain the given answer | A1 | [2] |
| Answer | Marks |
|---|---|
| Using the correct method for the moduli, divide scalar products by products of moduli and express \(\cos AOP = \cos BOP\) in terms of \(\lambda\), or in terms of \(\lambda\) and \(OP\) | M1* |
| Answer | Marks | Guidance |
|---|---|---|
| Using the correct method for the moduli, divide each expression by twice the product of the relevant moduli and express \(\cos AOP = \cos BOP\) in terms of \(\lambda\), or \(\lambda\) and \(OP\) | M1* | |
| Obtain a correct equation in any form, e.g. \(\frac{9 + 2\lambda}{3\sqrt{(9 + 4\lambda + 12\lambda^2)}} = \frac{11 + 14\lambda}{5\sqrt{(9 + 4\lambda + 12\lambda^2)}}\) | A1 | |
| Solve for \(\lambda\) | M1(dep*) | |
| Obtain \(\lambda = \frac{3}{8}\) | A1 | [5] |
| Answer | Marks |
|---|---|
| B1 | [1] |
**(i) Use a correct method to express $\overline{OP}$ in terms of $\lambda$**
Obtain the given answer | A1 | [2]
**(ii) EITHER: Use correct method to express scalar product of $\overline{OA}$ and $\overline{OP}$, or $\overline{OB}$ and $\overline{OP}$ in terms of $\lambda$**
Using the correct method for the moduli, divide scalar products by products of moduli and express $\cos AOP = \cos BOP$ in terms of $\lambda$, or in terms of $\lambda$ and $OP$ | M1* |
**OR1: Use correct method to express $OA^2 + OP^2 - AP^2$, or $OB^2 + OP^2 - BP^2$ in terms of $\lambda$**
Using the correct method for the moduli, divide each expression by twice the product of the relevant moduli and express $\cos AOP = \cos BOP$ in terms of $\lambda$, or $\lambda$ and $OP$ | M1* |
Obtain a correct equation in any form, e.g. $\frac{9 + 2\lambda}{3\sqrt{(9 + 4\lambda + 12\lambda^2)}} = \frac{11 + 14\lambda}{5\sqrt{(9 + 4\lambda + 12\lambda^2)}}$ | A1 |
Solve for $\lambda$ | M1(dep*) |
Obtain $\lambda = \frac{3}{8}$ | A1 | [5]
[SR: The M1* can also be earned by equating $\cos AOP$ or $\cos BOP$ to a sound attempt at $\cos \frac{1}{2}AOB$ and obtaining an equation in $\lambda$. The exact value of the cosine is $\sqrt{(13/15)}$, but accept non-exact working giving a value of $\lambda$ which rounds to 0.375, provided the spurious negative root of the quadratic in $\lambda$ is rejected.]
[SR: Allow a solution reaching $\lambda = \frac{3}{8}$ after cancelling identical incorrect expressions for $OP$ to score 4/5. The marking will run M1M1A0M1A1, or M1M1A1M1A0 in such cases.]
**(iii) Verify the given statement correctly**
| B1 | [1]7 With respect to the origin $O$, the position vectors of two points $A$ and $B$ are given by $\overrightarrow { O A } = \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k }$ and $\overrightarrow { O B } = 3 \mathbf { i } + 4 \mathbf { j }$. The point $P$ lies on the line through $A$ and $B$, and $\overrightarrow { A P } = \lambda \overrightarrow { A B }$.\\
(i) Show that $\overrightarrow { O P } = ( 1 + 2 \lambda ) \mathbf { i } + ( 2 + 2 \lambda ) \mathbf { j } + ( 2 - 2 \lambda ) \mathbf { k }$.\\
(ii) By equating expressions for $\cos A O P$ and $\cos B O P$ in terms of $\lambda$, find the value of $\lambda$ for which $O P$ bisects the angle $A O B$.\\
(iii) When $\lambda$ has this value, verify that $A P : P B = O A : O B$.
\hfill \mbox{\textit{CAIE P3 2011 Q7 [8]}}