| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Equation of plane containing line and point/parallel to vector |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths vectors question requiring standard techniques: finding plane equations from lines/points, using cross products for normals, calculating angles between planes, and finding perpendicular projections. While it involves several steps and Further Maths content (making it harder than typical A-level), each part uses routine methods without requiring novel insight or particularly complex reasoning. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-(-2\mathbf{i}-2\mathbf{j}+4\mathbf{k})+(-\mathbf{i}-2\mathbf{j}+\mathbf{k})=\mathbf{i}-3\mathbf{k}\) | B1 | OE. Finds direction vector from \(P\) to a point of \(l_1\) |
| \(\mathbf{r}=-2\mathbf{i}-2\mathbf{j}+4\mathbf{k}+\lambda(2\mathbf{i}-3\mathbf{j})+\mu(\mathbf{i}-3\mathbf{k})\) | B1 FT | OE. FT *their* \(\mathbf{i}-3\mathbf{k}\) |
| \(\mathbf{r}=-\mathbf{i}-2\mathbf{j}+\mathbf{k}+\lambda(2i-3j)+\mu(i-3k)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-3&0\\3&-1&3\end{vmatrix}=\begin{pmatrix}-9\\-6\\7\end{pmatrix}\) | M1 A1 | OE. Finds vector perpendicular to \(\Pi_2\) |
| \(-9(3)-6(0)+7(-2)=-41\) | M1 | Uses point on \(\Pi_2\) |
| \(9x+6y-7z=41\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-3&0\\1&0&-3\end{vmatrix}=\begin{pmatrix}9\\6\\3\end{pmatrix}\sim\begin{pmatrix}3\\2\\1\end{pmatrix}\) | M1 A1 | Finds vector perpendicular to \(\Pi_1\) |
| \(\begin{pmatrix}3\\2\\1\end{pmatrix}\cdot\begin{pmatrix}9\\6\\-7\end{pmatrix}=\sqrt{14}\sqrt{166}\cos\alpha\) leading to \(\cos\alpha=\dfrac{32}{\sqrt{14}\sqrt{166}}\) | DM1 A1 FT | Dot product using their normal vectors |
| \(48.4°\) | A1 | \(0.845\) rad |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{OF}=\overrightarrow{OQ}+\overrightarrow{QF}=-5\begin{pmatrix}-2\\-2\\4\end{pmatrix}+t\begin{pmatrix}9\\6\\-7\end{pmatrix}=\begin{pmatrix}10+9t\\10+6t\\-20-7t\end{pmatrix}\) | M1 A1 FT | Scales \(\overrightarrow{OP}\) by a factor of \(\pm5\) and uses multiple of their normal to \(\Pi_2\) |
| \(9(10+9t)+6(10+6t)-7(-20-7t)=41\) leading to \(290+166t=41\) | DM1 | Substitutes into the equation of \(\Pi_2\) |
| \(t=-\frac{3}{2}\) leading to \(\overrightarrow{OF}=\begin{pmatrix}-\frac{7}{2}\\1\\-\frac{19}{2}\end{pmatrix}\) | A1 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-(-2\mathbf{i}-2\mathbf{j}+4\mathbf{k})+(-\mathbf{i}-2\mathbf{j}+\mathbf{k})=\mathbf{i}-3\mathbf{k}$ | B1 | OE. Finds direction vector from $P$ to a point of $l_1$ |
| $\mathbf{r}=-2\mathbf{i}-2\mathbf{j}+4\mathbf{k}+\lambda(2\mathbf{i}-3\mathbf{j})+\mu(\mathbf{i}-3\mathbf{k})$ | B1 FT | OE. FT *their* $\mathbf{i}-3\mathbf{k}$ |
| $\mathbf{r}=-\mathbf{i}-2\mathbf{j}+\mathbf{k}+\lambda(2i-3j)+\mu(i-3k)$ | | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-3&0\\3&-1&3\end{vmatrix}=\begin{pmatrix}-9\\-6\\7\end{pmatrix}$ | M1 A1 | OE. Finds vector perpendicular to $\Pi_2$ |
| $-9(3)-6(0)+7(-2)=-41$ | M1 | Uses point on $\Pi_2$ |
| $9x+6y-7z=41$ | A1 | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-3&0\\1&0&-3\end{vmatrix}=\begin{pmatrix}9\\6\\3\end{pmatrix}\sim\begin{pmatrix}3\\2\\1\end{pmatrix}$ | M1 A1 | Finds vector perpendicular to $\Pi_1$ |
| $\begin{pmatrix}3\\2\\1\end{pmatrix}\cdot\begin{pmatrix}9\\6\\-7\end{pmatrix}=\sqrt{14}\sqrt{166}\cos\alpha$ leading to $\cos\alpha=\dfrac{32}{\sqrt{14}\sqrt{166}}$ | DM1 A1 FT | Dot product using their normal vectors |
| $48.4°$ | A1 | $0.845$ rad |
---
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{OF}=\overrightarrow{OQ}+\overrightarrow{QF}=-5\begin{pmatrix}-2\\-2\\4\end{pmatrix}+t\begin{pmatrix}9\\6\\-7\end{pmatrix}=\begin{pmatrix}10+9t\\10+6t\\-20-7t\end{pmatrix}$ | M1 A1 FT | Scales $\overrightarrow{OP}$ by a factor of $\pm5$ and uses multiple of their normal to $\Pi_2$ |
| $9(10+9t)+6(10+6t)-7(-20-7t)=41$ leading to $290+166t=41$ | DM1 | Substitutes into the equation of $\Pi_2$ |
| $t=-\frac{3}{2}$ leading to $\overrightarrow{OF}=\begin{pmatrix}-\frac{7}{2}\\1\\-\frac{19}{2}\end{pmatrix}$ | A1 | |
---6 The lines $l _ { 1 }$ and $l _ { 2 }$ have equations $\mathbf { r } = - \mathbf { i } - 2 \mathbf { j } + \mathbf { k } + s ( 2 \mathbf { i } - 3 \mathbf { j } )$ and $\mathbf { r } = 3 \mathbf { i } - 2 \mathbf { k } + t ( 3 \mathbf { i } - \mathbf { j } + 3 \mathbf { k } )$ respectively.
The plane $\Pi _ { 1 }$ contains $l _ { 1 }$ and the point $P$ with position vector $- 2 \mathbf { i } - 2 \mathbf { j } + 4 \mathbf { k }$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of $\Pi _ { 1 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b } + \mu \mathbf { c }$.\\
The plane $\Pi _ { 2 }$ contains $l _ { 2 }$ and is parallel to $l _ { 1 }$.
\item Find an equation of $\Pi _ { 2 }$, giving your answer in the form $\mathrm { ax } + \mathrm { by } + \mathrm { cz } = \mathrm { d }$.
\item Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
\item The point $Q$ is such that $\overrightarrow { \mathrm { OQ } } = - 5 \overrightarrow { \mathrm { OP } }$.
Find the position vector of the foot of the perpendicular from the point $Q$ to $\Pi _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q6 [15]}}