| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Line lies in or parallel to plane |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question testing routine concepts: checking if a line is parallel to a plane (dot product with normal = 0), finding when a line lies in a plane (substituting a point), and using the distance formula from point to plane. All techniques are textbook exercises with straightforward algebraic manipulation, making it slightly easier than average for Further Maths content. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Calculate scalar product of direction of \(l\) and normal to \(p\) | M1 | |
| Obtain \(4\times2 + 3\times(-2) + (-2)\times1 = 0\) and conclude accordingly | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \((a, 1, 4)\) in equation of \(p\) and solve for \(a\) | M1 | |
| Obtain \(a = 4\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: Attempt use of perpendicular distance formula using \((a, 1, 4)\) | M1 | |
| Obtain \(\frac{2a - 2 + 4 - 10}{\sqrt{4+4+1}} = 6\) | A1 | |
| Obtain \(a = 13\) | A1 | |
| Attempt solution of \(\frac{2a-8}{3} = -6\) | M1 | |
| Obtain \(a = -5\) | A1 | |
| Or: Form equation of parallel plane and substitute \((a, 1, 4)\) | M1 | |
| Obtain \(\frac{2a+2}{3} - \frac{10}{3} = 6\) | A1 | |
| Obtain \(a = 13\) | A1 | |
| Solve \(\frac{2a+2}{3} - \frac{10}{3} = -6\) | M1 | |
| Obtain \(a = -5\) | A1 | |
| Or: State vector from point on plane to \((a, 1, 4)\) e.g. \(\begin{pmatrix}a-5\\1\\4\end{pmatrix}\) or \(\begin{pmatrix}a\\1\\-6\end{pmatrix}\) | B1 | |
| Calculate component in direction of unit normal and equate to 6: \(\frac{1}{3}\begin{pmatrix}a-5\\1\\4\end{pmatrix}\cdot\begin{pmatrix}2\\-2\\1\end{pmatrix} = 6\) | M1 | |
| Obtain \(a = 13\) | A1 | |
| Solve \(\frac{1}{3}\begin{pmatrix}a-5\\1\\4\end{pmatrix}\cdot\begin{pmatrix}2\\-2\\1\end{pmatrix} = -6\) | M1 | |
| Obtain \(a = -5\) | A1 | |
| Or: State perpendicular line \(\mathbf{r} = \begin{pmatrix}a\\1\\4\end{pmatrix} + \mu\begin{pmatrix}2\\-2\\1\end{pmatrix}\) | B1 | |
| Substitute components for \(p\) and solve for \(\mu\) | M1 | |
| Obtain \(\mu = \frac{8-2a}{9}\) | A1 | |
| Equate distance between \((a, 1, 4)\) and foot of perpendicular to \(\pm 6\) | M1 | |
| Obtain \(\frac{3(8-2a)}{9} = \pm 6\) and hence \(-5\) and \(13\) | A1 | [5] |
## Question 9:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Calculate scalar product of direction of $l$ and normal to $p$ | M1 | |
| Obtain $4\times2 + 3\times(-2) + (-2)\times1 = 0$ and conclude accordingly | A1 | [2] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $(a, 1, 4)$ in equation of $p$ and solve for $a$ | M1 | |
| Obtain $a = 4$ | A1 | [2] |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** Attempt use of perpendicular distance formula using $(a, 1, 4)$ | M1 | |
| Obtain $\frac{2a - 2 + 4 - 10}{\sqrt{4+4+1}} = 6$ | A1 | |
| Obtain $a = 13$ | A1 | |
| Attempt solution of $\frac{2a-8}{3} = -6$ | M1 | |
| Obtain $a = -5$ | A1 | |
| **Or:** Form equation of parallel plane and substitute $(a, 1, 4)$ | M1 | |
| Obtain $\frac{2a+2}{3} - \frac{10}{3} = 6$ | A1 | |
| Obtain $a = 13$ | A1 | |
| Solve $\frac{2a+2}{3} - \frac{10}{3} = -6$ | M1 | |
| Obtain $a = -5$ | A1 | |
| **Or:** State vector from point on plane to $(a, 1, 4)$ e.g. $\begin{pmatrix}a-5\\1\\4\end{pmatrix}$ or $\begin{pmatrix}a\\1\\-6\end{pmatrix}$ | B1 | |
| Calculate component in direction of unit normal and equate to 6: $\frac{1}{3}\begin{pmatrix}a-5\\1\\4\end{pmatrix}\cdot\begin{pmatrix}2\\-2\\1\end{pmatrix} = 6$ | M1 | |
| Obtain $a = 13$ | A1 | |
| Solve $\frac{1}{3}\begin{pmatrix}a-5\\1\\4\end{pmatrix}\cdot\begin{pmatrix}2\\-2\\1\end{pmatrix} = -6$ | M1 | |
| Obtain $a = -5$ | A1 | |
| **Or:** State perpendicular line $\mathbf{r} = \begin{pmatrix}a\\1\\4\end{pmatrix} + \mu\begin{pmatrix}2\\-2\\1\end{pmatrix}$ | B1 | |
| Substitute components for $p$ and solve for $\mu$ | M1 | |
| Obtain $\mu = \frac{8-2a}{9}$ | A1 | |
| Equate distance between $(a, 1, 4)$ and foot of perpendicular to $\pm 6$ | M1 | |
| Obtain $\frac{3(8-2a)}{9} = \pm 6$ and hence $-5$ and $13$ | A1 | [5] |
---9 The line $l$ has equation $\mathbf { r } = \left( \begin{array} { l } a \\ 1 \\ 4 \end{array} \right) + \lambda \left( \begin{array} { r } 4 \\ 3 \\ - 2 \end{array} \right)$, where $a$ is a constant. The plane $p$ has equation $2 x - 2 y + z = 10$.\\
(i) Given that $l$ does not lie in $p$, show that $l$ is parallel to $p$.\\
(ii) Find the value of $a$ for which $l$ lies in $p$.\\
(iii) It is now given that the distance between $l$ and $p$ is 6 . Find the possible values of $a$.
\hfill \mbox{\textit{CAIE P3 2011 Q9 [9]}}