| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.3 This is a straightforward two-part vectors question requiring standard techniques: (a) finding angle between vectors using dot product formula, and (b) finding a point equidistant from two given points by setting up and solving |AP|=|BP|. Both parts are routine applications of A-level vector methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(\overrightarrow{AB} = \begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix}\) | B1 | OE. Allow \(\pm\) |
| Use the correct process to calculate the scalar product of a pair of relevant vectors, e.g. \(\overrightarrow{AB}\) and a direction vector for \(l\) | M1 | \((2+2-3=1)\) |
| Using the correct process for the moduli, divide the scalar product by the product of the moduli of the two vectors and evaluate the inverse cosine of the result | M1 | \(\cos^{-1}\left(\frac{1}{\sqrt{6}\sqrt{14}}\right)\) |
| Obtain answer \(83.7°\) or \(1.46\) radians | A1 | Or answers rounding to \(83.7°\) or \(1.46\) radians |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(\pm\overrightarrow{AP}\) and \(\pm\overrightarrow{BP}\) in component form, i.e. \((1+\lambda, 1-2\lambda, \lambda)\) and \((-1+\lambda, 2-2\lambda, 3+\lambda)\), or equivalent | B1 | |
| Form an equation in \(\lambda\) by equating moduli or by using \(\cos BAP = \cos ABP\) | *M1 | |
| Obtain a correct equation in any form \((1+\lambda)^2 + (1-2\lambda)^2 + \lambda^2 = (\lambda-1)^2 + (2-2\lambda)^2 + (\lambda+3)^2\) | A1 | Or \((1+\lambda)\sqrt{14-4\lambda+6\lambda^2} = (13-\lambda)\sqrt{2-2\lambda+6\lambda^2}\) \((83\lambda^3 - 528\lambda^2 + 207\lambda - 162 = 0)\) |
| Solve for \(\lambda\) and obtain position vector | DM1 | \([\lambda = 6]\) |
| Obtain correct position vector for \(P\) in any form, e.g. \((8, -9, 7)\) or \(8\mathbf{i} - 9\mathbf{j} + 7\mathbf{k}\) | A1 | Accept coordinates |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $\overrightarrow{AB} = \begin{pmatrix} 2 \\ -1 \\ -3 \end{pmatrix}$ | B1 | OE. Allow $\pm$ |
| Use the correct process to calculate the scalar product of a pair of relevant vectors, e.g. $\overrightarrow{AB}$ and a direction vector for $l$ | M1 | $(2+2-3=1)$ |
| Using the correct process for the moduli, divide the scalar product by the product of the moduli of the two vectors and evaluate the inverse cosine of the result | M1 | $\cos^{-1}\left(\frac{1}{\sqrt{6}\sqrt{14}}\right)$ |
| Obtain answer $83.7°$ or $1.46$ radians | A1 | Or answers rounding to $83.7°$ or $1.46$ radians |
---
## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $\pm\overrightarrow{AP}$ and $\pm\overrightarrow{BP}$ in component form, i.e. $(1+\lambda, 1-2\lambda, \lambda)$ and $(-1+\lambda, 2-2\lambda, 3+\lambda)$, or equivalent | B1 | |
| Form an equation in $\lambda$ by equating moduli or by using $\cos BAP = \cos ABP$ | *M1 | |
| Obtain a correct equation in any form $(1+\lambda)^2 + (1-2\lambda)^2 + \lambda^2 = (\lambda-1)^2 + (2-2\lambda)^2 + (\lambda+3)^2$ | A1 | Or $(1+\lambda)\sqrt{14-4\lambda+6\lambda^2} = (13-\lambda)\sqrt{2-2\lambda+6\lambda^2}$ $(83\lambda^3 - 528\lambda^2 + 207\lambda - 162 = 0)$ |
| Solve for $\lambda$ and obtain position vector | DM1 | $[\lambda = 6]$ |
| Obtain correct position vector for $P$ in any form, e.g. $(8, -9, 7)$ or $8\mathbf{i} - 9\mathbf{j} + 7\mathbf{k}$ | A1 | Accept coordinates |
---8 With respect to the origin $O$, the points $A$ and $B$ have position vectors given by $\overrightarrow { O A } = \left( \begin{array} { l } 1 \\ 2 \\ 1 \end{array} \right)$ and $\overrightarrow { O B } = \left( \begin{array} { r } 3 \\ 1 \\ - 2 \end{array} \right)$. The line $l$ has equation $\mathbf { r } = \left( \begin{array} { l } 2 \\ 3 \\ 1 \end{array} \right) + \lambda \left( \begin{array} { r } 1 \\ - 2 \\ 1 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find the acute angle between the directions of $A B$ and $l$.
\item Find the position vector of the point $P$ on $l$ such that $A P = B P$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q8 [9]}}