Date | November 2013 | Marks available | 3 | Reference code | 13N.2.hl.TZ0.7 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Show that | Question number | 7 | Adapted from | N/A |
Question
The vectors a and b are such that a =(3cosθ+6)i +7 j and b =(cosθ−2)i +(1+sinθ)j.
Given that a and b are perpendicular,
show that 3sin2θ−7sinθ+2=0;
find the smallest possible positive value of θ.
Markscheme
attempting to form (3cosθ+6)(cosθ−2)+7(1+sinθ)=0 M1
3cos2θ−12+7sinθ+7=0 A1
3(1−sin2θ)+7sinθ−5=0 M1
3sin2θ−7sinθ+2=0 AG
[3 marks]
attempting to solve algebraically (including substitution) or graphically for sinθ (M1)
sinθ=13 (A1)
θ=0.340 (=19.5∘) A1
[3 marks]
Examiners report
Part (a) was very well done. Most candidates were able to use the scalar product and cos2θ=1−sin2θ to show the required result.
Part (b) was reasonably well done. A few candidates confused ‘smallest possible positive value’ with a minimum function value. Some candidates gave θ=0.34 as their final answer.