OCR C4 2006 June — Question 7 8 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2006
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeLine intersection: show lines are skew
DifficultyStandard +0.3 This is a standard C4 vectors question testing understanding of skew vs intersecting lines. Part (i) requires setting up equations for intersection and finding when they're inconsistent (routine algebraic manipulation). Part (ii) uses the special value of a to find the intersection point. The concepts are core syllabus material with straightforward execution, making it slightly easier than average.
Spec4.04b Plane equations: cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting

Two lines have vector equations $$\mathbf{r} = \mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + \lambda(3\mathbf{i} + \mathbf{j} + a\mathbf{k})$$ and $$\mathbf{r} = -8\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + \mu(\mathbf{i} - 2\mathbf{j} - \mathbf{k}),$$ where \(a\) is a constant.
  1. Given that the lines are skew, find the value that \(a\) cannot take. [6]
  2. Given instead that the lines intersect, find the point of intersection. [2]

(i)
AnswerMarks Guidance
Produce at least 2 of the 3 relevant eqns in \(\lambda\) and \(\mu\)M1 e.g. \(1 + 3\lambda = -8 + \mu\), \(-2 + \lambda = 2 - 2\mu\)
Solve the 2 eqns in \(\lambda\) & \(\mu\) as far as \(\lambda = \ldots\) or \(\mu = \ldots\)M1
1st solution: \(\lambda = -2\) or \(\mu = 3\)A1
2nd solution: \(\mu = 3\) or \(\lambda = -2\)A1√ f.t.
Substitute their \(\lambda\) and \(\mu\) into 3rd eqn and find 'a'M1
Obtain \(a = 2\) & clearly state that \(a\) cannot be 2A1 6
(ii)
AnswerMarks Guidance
Subst their \(\lambda\) or \(\mu\) (& poss a) into either line eqnM1
Point of intersection is \(-5i - 4j\)A1 2 Accept any format, No f.t. here
N.B. In this question, award marks irrespective of labelling of parts
### (i)
Produce at least 2 of the 3 relevant eqns in $\lambda$ and $\mu$ | M1 | e.g. $1 + 3\lambda = -8 + \mu$, $-2 + \lambda = 2 - 2\mu$

Solve the 2 eqns in $\lambda$ & $\mu$ as far as $\lambda = \ldots$ or $\mu = \ldots$ | M1 |

1st solution: $\lambda = -2$ or $\mu = 3$ | A1 |

2nd solution: $\mu = 3$ or $\lambda = -2$ | A1√ | f.t.

Substitute their $\lambda$ and $\mu$ into 3rd eqn and find 'a' | M1 |

Obtain $a = 2$ & clearly state that $a$ cannot be 2 | A1 | 6

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### (ii)
Subst their $\lambda$ or $\mu$ (& poss a) into either line eqn | M1 |
Point of intersection is $-5i - 4j$ | A1 | 2 Accept any format, No f.t. here

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N.B. In this question, award marks irrespective of labelling of parts

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Two lines have vector equations
$$\mathbf{r} = \mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + \lambda(3\mathbf{i} + \mathbf{j} + a\mathbf{k})$$ and $$\mathbf{r} = -8\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + \mu(\mathbf{i} - 2\mathbf{j} - \mathbf{k}),$$
where $a$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Given that the lines are skew, find the value that $a$ cannot take. [6]
\item Given instead that the lines intersect, find the point of intersection. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR C4 2006 Q7 [8]}}