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Date May 2008 Marks available 7 Reference code 08M.2.sl.TZ1.7
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 7 Adapted from N/A

Question

Let  \({\boldsymbol{v}} = 3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}}\) and \({\boldsymbol{w}} = {\boldsymbol{i}} + 2{\boldsymbol{j}} - 3{\boldsymbol{k}}\) . The vector \({\boldsymbol{v}} + p{\boldsymbol{w}}\) is perpendicular to w. Find the value of p.

Markscheme

\(p{\boldsymbol{w}} = p{\boldsymbol{i}} + 2p{\boldsymbol{j}} - 3p{\boldsymbol{k}}\) (seen anywhere)     (A1)

attempt to find \({\boldsymbol{v}} + p{\boldsymbol{w}}\)     (M1)

e.g. \(3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}} + p({\boldsymbol{i}} + 2{\boldsymbol{j}} - 3{\boldsymbol{k}})\)

collecting terms \((3 + p){\boldsymbol{i}} + (4 + 2p){\boldsymbol{j}} + (1 - 3p){\boldsymbol{k}}\)     A1

attempt to find the dot product     (M1)

e.g. \(1(3 + p) + 2(4 + 2p) - 3(1 - 3p)\)

setting their dot product equal to 0     (M1)

e.g. \(1(3 + p) + 2(4 + 2p) - 3(1 - 3p) = 0\)

simplifying     A1

e.g. \(3 + p + 8 + 4p - 3 + 9p = 0\) , \(14p + 8 = 0\)

\(p = - 0.571\) \(\left( { - \frac{8}{{14}}} \right)\)     A1     N3

[7 marks]

Examiners report

This question was very poorly done with many leaving it blank. Of those that did attempt it, most were able to find \({\boldsymbol{v}} + p{\boldsymbol{w}}\) but really did not know how to proceed from there. They tried many approaches, such as, finding magnitudes, using negative reciprocals, or calculating the angle between two vectors. A few had the idea that the scalar product should equal zero but had trouble trying to set it up.

Syllabus sections

Topic 4 - Vectors » 4.2 » Perpendicular vectors; parallel vectors.
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