OCR FP3 2009 January — Question 3 6 marks

Exam BoardOCR
ModuleFP3 (Further Pure Mathematics 3)
Year2009
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeShortest distance between two skew lines
DifficultyChallenging +1.2 This is a standard Further Maths question on skew lines requiring the cross product of direction vectors and the scalar triple product formula. While it involves multiple steps and vector manipulation beyond A-level Core, it's a textbook application of well-defined techniques with no novel insight required, making it moderately above average difficulty.
Spec4.04e Line intersections: parallel, skew, or intersecting4.04i Shortest distance: between a point and a line

3 Two skew lines have equations $$\frac { x } { 2 } = \frac { y + 3 } { 1 } = \frac { z - 6 } { 3 } \quad \text { and } \quad \frac { x - 5 } { 3 } = \frac { y + 1 } { 1 } = \frac { z - 7 } { 5 } .$$
  1. Find the direction of the common perpendicular to the lines.
  2. Find the shortest distance between the lines.

3 Two skew lines have equations

$$\frac { x } { 2 } = \frac { y + 3 } { 1 } = \frac { z - 6 } { 3 } \quad \text { and } \quad \frac { x - 5 } { 3 } = \frac { y + 1 } { 1 } = \frac { z - 7 } { 5 } .$$

(i) Find the direction of the common perpendicular to the lines.\\
(ii) Find the shortest distance between the lines.

\hfill \mbox{\textit{OCR FP3 2009 Q3 [6]}}