Date | November 2016 | Marks available | 4 | Reference code | 16N.1.SL.TZ0.S_4 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find | Question number | S_4 | Adapted from | N/A |
Question
The position vectors of points P and Q are i ++ 2 j −− k and 7i ++ 3j −− 4k respectively.
Find a vector equation of the line that passes through P and Q.
The line through P and Q is perpendicular to the vector 2i ++ nk. Find the value of nn.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
valid attempt to find direction vector (M1)
eg→PQ, →QP−−→PQ, −−→QP
correct direction vector (or multiple of) (A1)
eg6i ++ j −− 3k
any correct equation in the form r == a ++ tb (any parameter for tt) A2 N3
where a is i ++ 2j −− k or 7i ++ 3j −− 4k , and b is a scalar multiple of 6i ++ j −− 3k
egr == 7i ++ 3j −− 4k ++ t(6i ++ j −− 3k), r =(1+6s2+1s−1−3s), r=(12−1)+t(−6−13)=⎛⎜⎝1+6s2+1s−1−3s⎞⎟⎠, r=⎛⎜⎝12−1⎞⎟⎠+t⎛⎜⎝−6−13⎞⎟⎠
Notes: Award A1 for the form a ++ tb, A1 for the form L == a ++ tb, A0 for the form r == b ++ ta.
[4 marks]
correct expression for scalar product (A1)
eg6×2+1×0+(−3)×n, −3n+126×2+1×0+(−3)×n, −3n+12
setting scalar product equal to zero (seen anywhere) (M1)
egu ∙ v =0, −3n+12=0
n=4 A1 N2
[3 marks]