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.