| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to line |
| Difficulty | Standard +0.3 This is a standard three-part vectors question requiring routine techniques: finding angle between vectors using dot product, finding foot of perpendicular using scalar parameter, and using midpoint formula for reflection. All methods are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using the correct process find the scalar product of direction vectors of \(l\) and \(\overrightarrow{OA}\) | M1 | \((1,5,6)\cdot(-1,2,3)=-1+5.2+6.3=-1+10+18\) |
| Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result | M1 | Their scalar product \(\div[\sqrt{(1^2+5^2+6^2)}\sqrt{((-1)^2+2^2+3^2)}]\). Angle \(=\cos^{-1}\frac{27}{\sqrt{62}\sqrt{14}}\) |
| Obtain answer \(23.6°\) | A1 | AWRT \(23.6°\). \(23.5889°\). Radians \(0.412\) scores A0 \((0.4117...)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking a general point \(P\) on \(l\), state \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) in component form, e.g. \((3-\lambda,-5+2\lambda,-5+3\lambda)\) | B1 | Note: \((4,1,0)\) or \((4,1,1)\), for \(4\mathbf{i}+\mathbf{k}\) is not MR, but M1 possible |
| Either equate scalar product of \(\overrightarrow{AP}\) and direction vector of \(l\) to zero and solve for \(\lambda\), or use Pythagoras in a relevant triangle and solve for \(\lambda\) | M1 | \((3-\lambda,-5+2\lambda,-5+3\lambda)\cdot(-1,2,3)=0\); \(-3-10-15+\lambda+4\lambda+9\lambda=0\) |
| Obtain \(\lambda=2\) | A1 | \(\lambda=2\) |
| State that the position vector \(\overrightarrow{OP^*}\) of the foot is \(2\mathbf{i}+4\mathbf{j}+7\mathbf{k}\) | A1 | OE. Condone coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Set up a correct method for finding the position vector of the reflection of \(A\) in \(l\) | M1 | For all methods, allow a sign error in one component only: \(\overrightarrow{OA'}=\overrightarrow{OP^*}+(\overrightarrow{OP^*}-\overrightarrow{OA})\); or \(\overrightarrow{OA'}=\overrightarrow{OP^*}-(\overrightarrow{OA}-\overrightarrow{OP^*})\); or \(\overrightarrow{OA'}=\overrightarrow{OA}+2(\overrightarrow{OP^*}-\overrightarrow{OA})\); or midpoint \(\overrightarrow{OP^*}=(\overrightarrow{OA}+\overrightarrow{OA'})/2\) with their \(\lambda\) value substituted |
| Obtain answer \(3\mathbf{i}+3\mathbf{j}+8\mathbf{k}\) or \(3\left(\mathbf{i}+\mathbf{j}+\frac{8}{3}\right)\) | A1 | OE. Condone coordinates \(x=3,y=3,z=8\). No method shown and correct answer 2/2 |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using the correct process find the scalar product of direction vectors of $l$ and $\overrightarrow{OA}$ | M1 | $(1,5,6)\cdot(-1,2,3)=-1+5.2+6.3=-1+10+18$ |
| Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result | M1 | Their scalar product $\div[\sqrt{(1^2+5^2+6^2)}\sqrt{((-1)^2+2^2+3^2)}]$. Angle $=\cos^{-1}\frac{27}{\sqrt{62}\sqrt{14}}$ |
| Obtain answer $23.6°$ | A1 | AWRT $23.6°$. $23.5889°$. Radians $0.412$ scores A0 $(0.4117...)$ |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking a general point $P$ on $l$, state $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) in component form, e.g. $(3-\lambda,-5+2\lambda,-5+3\lambda)$ | B1 | Note: $(4,1,0)$ or $(4,1,1)$, for $4\mathbf{i}+\mathbf{k}$ is not MR, but M1 possible |
| Either equate scalar product of $\overrightarrow{AP}$ and direction vector of $l$ to zero and solve for $\lambda$, or use Pythagoras in a relevant triangle and solve for $\lambda$ | M1 | $(3-\lambda,-5+2\lambda,-5+3\lambda)\cdot(-1,2,3)=0$; $-3-10-15+\lambda+4\lambda+9\lambda=0$ |
| Obtain $\lambda=2$ | A1 | $\lambda=2$ |
| State that the position vector $\overrightarrow{OP^*}$ of the foot is $2\mathbf{i}+4\mathbf{j}+7\mathbf{k}$ | A1 | OE. Condone coordinates |
---
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Set up a correct method for finding the position vector of the reflection of $A$ in $l$ | M1 | For all methods, allow a sign error in one component only: $\overrightarrow{OA'}=\overrightarrow{OP^*}+(\overrightarrow{OP^*}-\overrightarrow{OA})$; or $\overrightarrow{OA'}=\overrightarrow{OP^*}-(\overrightarrow{OA}-\overrightarrow{OP^*})$; or $\overrightarrow{OA'}=\overrightarrow{OA}+2(\overrightarrow{OP^*}-\overrightarrow{OA})$; or midpoint $\overrightarrow{OP^*}=(\overrightarrow{OA}+\overrightarrow{OA'})/2$ with their $\lambda$ value substituted |
| Obtain answer $3\mathbf{i}+3\mathbf{j}+8\mathbf{k}$ or $3\left(\mathbf{i}+\mathbf{j}+\frac{8}{3}\right)$ | A1 | OE. Condone coordinates $x=3,y=3,z=8$. No method shown and correct answer 2/2 |9 With respect to the origin $O$, the point $A$ has position vector given by $\overrightarrow { O A } = \mathbf { i } + 5 \mathbf { j } + 6 \mathbf { k }$. The line $l$ has vector equation $\mathbf { r } = 4 \mathbf { i } + \mathbf { k } + \lambda ( - \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } )$.
\begin{enumerate}[label=(\alph*)]
\item Find in degrees the acute angle between the directions of $O A$ and $l$.
\item Find the position vector of the foot of the perpendicular from $A$ to $l$.
\item Hence find the position vector of the reflection of $A$ in $l$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q9 [9]}}