OCR MEI Further Pure Core 2023 June — Question 16 10 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2023
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeTrigonometric substitution to simplify integral
DifficultyChallenging +1.8 This is a challenging 3D coordinate geometry problem requiring multiple sophisticated techniques: calculating distance from point to plane using the perpendicular distance formula, finding distance from point to line (requiring projection or perpendicular foot), then solving the resulting equation. While each component uses standard Further Maths methods, the multi-step nature, algebraic complexity of equating two distance expressions, and the need to handle the parameter b systematically make this substantially harder than typical A-level questions but not exceptionally difficult for Further Maths students.
Spec4.04j Shortest distance: between a point and a plane

16 The point \(P ( 4,1,0 )\) is equidistant from the plane \(2 x + y + 2 z = 0\) and the line \(\frac { x - 3 } { 2 } = \frac { y - 1 } { b } = \frac { z + 5 } { 3 }\), where \(b > 0\). Determine the value of \(b\).

16 The point $P ( 4,1,0 )$ is equidistant from the plane $2 x + y + 2 z = 0$ and the line $\frac { x - 3 } { 2 } = \frac { y - 1 } { b } = \frac { z + 5 } { 3 }$, where $b > 0$.

Determine the value of $b$.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q16 [10]}}