| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to plane |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring: (i) finding intersection of skew-looking lines by solving simultaneous equations, (ii) finding a plane equation from two lines then computing point-to-plane distance using the standard formula, and (iii) point-to-line distance using cross product or projection. While each technique is standard for FM students, the combination of three distinct vector geometry methods in one question, plus the algebraic manipulation involved, places this moderately above average difficulty. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10g Problem solving with vectors: in geometry4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04g Vector product: a x b perpendicular vector4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solve two equations: \(9+2\lambda = 7+2\mu\) and \(13+3\lambda = -2-3\mu\) | M1 | |
| to obtain \(\lambda = -3\) and \(\mu = -2\) | A1 | |
| Use third equation to obtain \(a = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Normal to the plane is \(\mathbf{n} = -12\mathbf{i} + 4\mathbf{k}\) | M1 A1 | |
| Perpendicular distance \(= \frac{1}{\sqrt{160}}\begin{pmatrix}-6\\6\\-8\end{pmatrix}\cdot\begin{pmatrix}-12\\0\\4\end{pmatrix} = \sqrt{10} = 3.16\) | M1 A1 | Alt method: Finds equation of plane M1, A1; Finds foot of perpendicular from P to plane M1; Hence length A1; Alt: Find equation of plane \(3x-z+7=0\) M1, A1; Use formula \(\frac{ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Cross product of direction of \(\mathbf{P}\) to \(l_2\) with direction of \(l_2\): \(\begin{pmatrix}-6\\6\\-8\end{pmatrix}\times\begin{pmatrix}-1\\2\\-3\end{pmatrix} = \begin{pmatrix}-2\\-10\\-6\end{pmatrix}\) | M1 A1 | Alt method: Find N (foot of perpendicular) in terms of parameter and uses scalar product with \(\mathbf{n}\) to find parameter M1, A1 so \(PN = \begin{pmatrix}3\\0\\-1\end{pmatrix}\) |
| Perpendicular distance from \(\mathbf{P}\) to \(l_2\) is \(\frac{ | 2\mathbf{i}+10\mathbf{j}+6\mathbf{k} | }{ |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve two equations: $9+2\lambda = 7+2\mu$ and $13+3\lambda = -2-3\mu$ | M1 | |
| to obtain $\lambda = -3$ and $\mu = -2$ | A1 | |
| Use third equation to obtain $a = 2$ | A1 | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Normal to the plane is $\mathbf{n} = -12\mathbf{i} + 4\mathbf{k}$ | M1 A1 | |
| Perpendicular distance $= \frac{1}{\sqrt{160}}\begin{pmatrix}-6\\6\\-8\end{pmatrix}\cdot\begin{pmatrix}-12\\0\\4\end{pmatrix} = \sqrt{10} = 3.16$ | M1 A1 | Alt method: Finds equation of plane M1, A1; Finds foot of perpendicular from P to plane M1; Hence length A1; Alt: Find equation of plane $3x-z+7=0$ M1, A1; Use formula $\frac{|3\times3+0\times1-6+7|}{\sqrt{9+1}} = \sqrt{10}$ M1, A1 |
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cross product of direction of $\mathbf{P}$ to $l_2$ with direction of $l_2$: $\begin{pmatrix}-6\\6\\-8\end{pmatrix}\times\begin{pmatrix}-1\\2\\-3\end{pmatrix} = \begin{pmatrix}-2\\-10\\-6\end{pmatrix}$ | M1 A1 | Alt method: Find N (foot of perpendicular) in terms of parameter and uses scalar product with $\mathbf{n}$ to find parameter M1, A1 so $PN = \begin{pmatrix}3\\0\\-1\end{pmatrix}$ |
| Perpendicular distance from $\mathbf{P}$ to $l_2$ is $\frac{|2\mathbf{i}+10\mathbf{j}+6\mathbf{k}|}{|-\mathbf{i}+2\mathbf{j}-3\mathbf{k}|} = \sqrt{10} = 3.16$ | M1 A1 | Find length PN M1, A1 |7 The lines $l _ { 1 }$ and $l _ { 2 }$ have vector equations
$$\mathbf { r } = a \mathbf { i } + 9 \mathbf { j } + 13 \mathbf { k } + \lambda ( \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = - 3 \mathbf { i } + 7 \mathbf { j } - 2 \mathbf { k } + \mu ( - \mathbf { i } + 2 \mathbf { j } - 3 \mathbf { k } )$$
respectively. It is given that $l _ { 1 }$ and $l _ { 2 }$ intersect.\\
(i) Find the value of the constant $a$.\\
The point $P$ has position vector $3 \mathbf { i } + \mathbf { j } + 6 \mathbf { k }$.\\
(ii) Find the perpendicular distance from $P$ to the plane containing $l _ { 1 }$ and $l _ { 2 }$.\\
(iii) Find the perpendicular distance from $P$ to $l _ { 2 }$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q7 [11]}}