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Date May 2008 Marks available 6 Reference code 08M.2.hl.TZ1.5
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 5 Adapted from N/A

Question

Find the vector equation of the line of intersection of the three planes represented by the following system of equations.

\[2x - 7y + 5z = 1\]

\[6x + 3y - z = - 1\]

\[ - 14x - 23y + 13z = 5\]

Markscheme

METHOD 1

(from GDC)

\(\left( {\begin{array}{*{20}{ccc|c}}
  1&0&{\frac{1}{6}}&{ - \frac{1}{{12}}} \\
  0&1&{ - \frac{2}{3}}&{ - \frac{1}{6}} \\
  0&0&0&0
\end{array}} \right)\)     (M1)

\(x + \frac{1}{6}\lambda = - \frac{1}{{12}}\)     A1

\(y - \frac{2}{3}\lambda = - \frac{1}{6}\)     A1

\(\boldsymbol{r} = \left( { - \frac{1}{{12}}\boldsymbol{i} - \frac{1}{6}\boldsymbol{j}} \right) + \lambda \left( { - \frac{1}{6}\boldsymbol{i} + \frac{2}{3}\boldsymbol{j} + \boldsymbol{k}} \right)\)     A1A1A1     N3

[6 marks] 

METHOD 2

(Elimination method either for equations or row reduction of matrix)

Eliminating one of the variables     M1A1

Finding a point on the line     (M1)A1

Finding the direction of the line     M1

The vector equation of the line     A1     N3

[6 marks]

Examiners report

A large number of candidates did not use their GDC in this question. Some candidates who attempted analytical solutions looked for a point solution although the question specifically states that the planes intersect in a line. Other candidates eliminated one variable and then had no clear strategy for proceeding with the solution.

Some candidates failed to write ‘r =’, and others did not give the equation in vector form.

Syllabus sections

Topic 4 - Core: Vectors » 4.7 » Intersections of: a line with a plane; two planes; three planes.

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