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 \( + \) j \( + \) 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.