CAIE P3 2012 November — Question 8 10 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2012
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with line
DifficultyStandard +0.3 This is a standard two-part vectors question requiring equating components to find intersection (routine algebraic manipulation), then finding a plane equation using cross product of direction vectors. While it involves multiple steps, all techniques are textbook-standard with no novel insight required, making it slightly easier than average.
Spec4.04b Plane equations: cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04f Line-plane intersection: find point

8 Two lines have equations $$\mathbf { r } = \left( \begin{array} { r } 5 \\ 1 \\ - 4 \end{array} \right) + s \left( \begin{array} { r } 1 \\ - 1 \\ 3 \end{array} \right) \quad \text { and } \quad \mathbf { r } = \left( \begin{array} { r } p \\ 4 \\ - 2 \end{array} \right) + t \left( \begin{array} { r } 2 \\ 5 \\ - 4 \end{array} \right) ,$$ where \(p\) is a constant. It is given that the lines intersect.
  1. Find the value of \(p\) and determine the coordinates of the point of intersection.
  2. Find the equation of the plane containing the two lines, giving your answer in the form \(a x + b y + c z = d\), where \(a , b , c\) and \(d\) are integers.

Question 8:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
State or imply general point of either line has coordinates \((5+s,\;1-s,\;-4+3s)\) or \((p+2t,\;4+5t,\;-2-4t)\)B1
Solve simultaneous equations and find \(s\) and \(t\)M1
Obtain \(s = 2\) and \(t = -1\) or equivalent in terms of \(p\)A1
Substitute in third equation to find \(p = 9\)A1
State point of intersection is \((7,\,-1,\,2)\)A1 [5]
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Either: Use scalar product to obtain a relevant equation in \(a, b, c\), e.g. \(a - b + 3c = 0\) or \(2a + 5b - 4c = 0\)M1
State two correct equations in \(a, b, c\)A1
Solve simultaneous equations to obtain at least one ratioDM1
Obtain \(a : b : c = -11 : 10 : 7\) or equivalentA1
Obtain equation \(-11x + 10y + 7z = -73\) or equivalent with integer coefficientsA1
Or 1: Calculate vector product of \(\begin{pmatrix}1\\-1\\3\end{pmatrix}\) and \(\begin{pmatrix}2\\5\\-4\end{pmatrix}\)M1
Obtain two correct components of the productA1
Obtain correct \(\begin{pmatrix}-11\\10\\7\end{pmatrix}\) or equivalentA1
Substitute coordinates of a relevant point in \(\mathbf{r}\cdot\mathbf{n} = d\) to find \(d\)DM1
Obtain equation \(-11x + 10y + 7z = -73\) or equivalent with integer coefficientsA1
Or 2: Using relevant vectors, form correctly a two-parameter equation for the planeM1
Obtain \(\mathbf{r} = \begin{pmatrix}5\\1\\-4\end{pmatrix} + \lambda\begin{pmatrix}1\\-1\\3\end{pmatrix} + \mu\begin{pmatrix}2\\5\\-4\end{pmatrix}\) or equivalentA1
State three equations in \(x, y, z, \lambda, \mu\)A1
Eliminate \(\lambda\) and \(\mu\)DM1
Obtain \(11x - 10y - 7z = 73\) or equivalent with integer coefficientsA1 [5]
## Question 8:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply general point of either line has coordinates $(5+s,\;1-s,\;-4+3s)$ or $(p+2t,\;4+5t,\;-2-4t)$ | B1 | |
| Solve simultaneous equations and find $s$ and $t$ | M1 | |
| Obtain $s = 2$ and $t = -1$ or equivalent in terms of $p$ | A1 | |
| Substitute in third equation to find $p = 9$ | A1 | |
| State point of intersection is $(7,\,-1,\,2)$ | A1 | [5] |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** Use scalar product to obtain a relevant equation in $a, b, c$, e.g. $a - b + 3c = 0$ or $2a + 5b - 4c = 0$ | M1 | |
| State two correct equations in $a, b, c$ | A1 | |
| Solve simultaneous equations to obtain at least one ratio | DM1 | |
| Obtain $a : b : c = -11 : 10 : 7$ or equivalent | A1 | |
| Obtain equation $-11x + 10y + 7z = -73$ or equivalent with integer coefficients | A1 | |
| **Or 1:** Calculate vector product of $\begin{pmatrix}1\\-1\\3\end{pmatrix}$ and $\begin{pmatrix}2\\5\\-4\end{pmatrix}$ | M1 | |
| Obtain two correct components of the product | A1 | |
| Obtain correct $\begin{pmatrix}-11\\10\\7\end{pmatrix}$ or equivalent | A1 | |
| Substitute coordinates of a relevant point in $\mathbf{r}\cdot\mathbf{n} = d$ to find $d$ | DM1 | |
| Obtain equation $-11x + 10y + 7z = -73$ or equivalent with integer coefficients | A1 | |
| **Or 2:** Using relevant vectors, form correctly a two-parameter equation for the plane | M1 | |
| Obtain $\mathbf{r} = \begin{pmatrix}5\\1\\-4\end{pmatrix} + \lambda\begin{pmatrix}1\\-1\\3\end{pmatrix} + \mu\begin{pmatrix}2\\5\\-4\end{pmatrix}$ or equivalent | A1 | |
| State three equations in $x, y, z, \lambda, \mu$ | A1 | |
| Eliminate $\lambda$ and $\mu$ | DM1 | |
| Obtain $11x - 10y - 7z = 73$ or equivalent with integer coefficients | A1 | [5] |
8 Two lines have equations

$$\mathbf { r } = \left( \begin{array} { r } 
5 \\
1 \\
- 4
\end{array} \right) + s \left( \begin{array} { r } 
1 \\
- 1 \\
3
\end{array} \right) \quad \text { and } \quad \mathbf { r } = \left( \begin{array} { r } 
p \\
4 \\
- 2
\end{array} \right) + t \left( \begin{array} { r } 
2 \\
5 \\
- 4
\end{array} \right) ,$$

where $p$ is a constant. It is given that the lines intersect.\\
(i) Find the value of $p$ and determine the coordinates of the point of intersection.\\
(ii) Find the equation of the plane containing the two lines, giving your answer in the form $a x + b y + c z = d$, where $a , b , c$ and $d$ are integers.

\hfill \mbox{\textit{CAIE P3 2012 Q8 [10]}}