OCR MEI C4 — Question 2 4 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeVector equation of a line
DifficultyModerate -0.3 This is a straightforward substitution problem requiring students to substitute the parametric equations into the plane equation, solve for λ, then find the point. It's slightly easier than average as it involves only routine algebraic manipulation with no conceptual challenges or multi-step problem-solving.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04f Line-plane intersection: find point
2 Find where the line \(\mathbf { r } = \left( \begin{array} { l } 1 \\ 2 \\ 0 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 3 \\ 2 \end{array} \right)\) meets the plane \(2 x + 3 y - 4 z - 5 = 0\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(x = 1+\lambda,\ y = 2+3\lambda,\ z = 2\lambda\) (say)B1
\(\Rightarrow 0 = 2(1+\lambda) + 3(2+3\lambda) - 4(2\lambda) - 5\)M1
\(\Rightarrow \lambda = -1\)A1
Required point is \((0, -1, -2)\)A1
Total: 4 
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 1+\lambda,\ y = 2+3\lambda,\ z = 2\lambda$ (say) | B1 | |
| $\Rightarrow 0 = 2(1+\lambda) + 3(2+3\lambda) - 4(2\lambda) - 5$ | M1 | |
| $\Rightarrow \lambda = -1$ | A1 | |
| Required point is $(0, -1, -2)$ | A1 | |
| **Total: 4** | | |

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2 Find where the line $\mathbf { r } = \left( \begin{array} { l } 1 \\ 2 \\ 0 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 3 \\ 2 \end{array} \right)$ meets the plane $2 x + 3 y - 4 z - 5 = 0$.

\hfill \mbox{\textit{OCR MEI C4  Q2 [4]}}
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Q1 4 Q2 4 Q3 6 Q4 7 Q5 5 Q6 6 Q7 4 Q8 19 Q9 17