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Date May 2016 Marks available 7 Reference code 16M.1.sl.TZ2.7
Level SL only Paper 1 Time zone TZ2
Command term Copy and complete and Find Question number 7 Adapted from N/A

Question

Let u \( =  - 3\)i \( + \) \( + \) k and v \( = m\)j \( + {\text{ }}n\)k , where \(m,{\text{ }}n \in \mathbb{R}\). Given that v is a unit vector perpendicular to u, find the possible values of \(m\) and of \(n\).

Markscheme

correct scalar product     (A1)

eg\(\,\,\,\,\,\)\(m + n\)

setting up their scalar product equal to 0 (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)u \( \bullet \) v \( = 0,{\text{ }} - 3(0) + 1(m) + 1(n) = 0,{\text{ }}m =  - n\)

correct interpretation of unit vector     (A1)

eg\(\,\,\,\,\,\)\(\sqrt {{0^2} + {m^2} + {n^2}}  = 1,{\text{ }}{m^2} + {n^2} = 1\)

valid attempt to solve their equations (must be in one variable)     M1

eg\(\,\,\,\,\,\)\({( - n)^2} + {n^2} = 1,{\text{ }}\sqrt {1 - {n^2}}  + n = 0,{\text{ }}{m^2} + {( - m)^2} = 1,{\text{ }}m - \sqrt {1 - {m^2}}  = 0\)

correct working     A1

eg\(\,\,\,\,\,\)\(2{n^2} = 1,{\text{ }}2{m^2} = 1,{\text{ }}\sqrt 2  = \frac{1}{n},{\text{ }}m =  \pm \frac{1}{{\sqrt 2 }}\)

both correct pairs     A2     N3

eg\(\,\,\,\,\,\)\(m = \frac{1}{{\sqrt 2 }}\) and \(n =  - \frac{1}{{\sqrt 2 }},{\text{ }}m =  - \frac{1}{{\sqrt 2 }}\) and \(n = \frac{1}{{\sqrt 2 }}\),

\(m = {(0.5)^{\frac{1}{2}}}\) and \(n =  - {(0.5)^{\frac{1}{2}}},{\text{ }}m =  - \sqrt {\frac{1}{2}} \) and \(n = \sqrt {\frac{1}{2}} \)

Note:     Award A0 for \(m =  \pm \frac{1}{{\sqrt 2 }},{\text{ }}n =  \pm \frac{1}{{\sqrt 2 }}\), or any other answer that does not clearly indicate the correct pairs.

[7 marks]

Examiners report

Most of the candidates recognized that the scalar product of the vectors must be zero. However, some did not find the correct scalar product because they did not multiply the correct corresponding vector components of u and v. In addition, the majority of candidates did not attempt to use the fact that the unit vector v has a magnitude of 1. For the small number of candidates who were successful in solving for \(m\) and/or \(n\), some did not clearly present the correct pairs of answers.

Syllabus sections

Topic 4 - Vectors » 4.2 » The scalar product of two vectors.
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