Date | May 2018 | Marks available | 1 | Reference code | 18M.2.hl.TZ1.11 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 11 | Adapted from | N/A |
Question
Two submarines A and B have their routes planned so that their positions at time t hours, 0 ≤ t < 20 , would be defined by the position vectors rA =⎛⎝24−1⎞⎠+t⎛⎝−11−0.15⎞⎠ and rB =⎛⎝03.2−2⎞⎠+t⎛⎝−0.51.20.1⎞⎠ relative to a fixed point on the surface of the ocean (all lengths are in kilometres).
To avoid the collision submarine B adjusts its velocity so that its position vector is now given by
rB =⎛⎝03.2−2⎞⎠+t⎛⎝−0.451.080.09⎞⎠.
Show that the two submarines would collide at a point P and write down the coordinates of P.
Show that submarine B travels in the same direction as originally planned.
Find the value of t when submarine B passes through P.
Find an expression for the distance between the two submarines in terms of t.
Find the value of t when the two submarines are closest together.
Find the distance between the two submarines at this time.
Markscheme
rA = rB (M1)
2 − t = − 0.5t ⇒ t = 4 A1
checking t = 4 satisfies 4 + t = 3.2 + 1.2t and − 1 − 0.15t = − 2 + 0.1t R1
P(−2, 8, −1.6) A1
Note: Do not award final A1 if answer given as column vector.
[4 marks]
0.9×⎛⎝−0.51.20.1⎞⎠=⎛⎝−0.451.080.09⎞⎠ A1
Note: Accept use of cross product equalling zero.
hence in the same direction AG
[1 mark]
⎛⎝−0.45t3.2+1.08t−2+0.09t⎞⎠=⎛⎝−28−1.6⎞⎠ M1
Note: The M1 can be awarded for any one of the resultant equations.
⇒t=409=4.44… A1
[2 marks]
rA − rB = ⎛⎝2−t4+t−1−0.15t⎞⎠−⎛⎝−0.45t3.2+1.08t−2+0.09t⎞⎠ (M1)(A1)
=⎛⎝2−0.55t0.8−0.08t1−0.24t⎞⎠ (A1)
Note: Accept rA − rB.
distance D=√(2−0.55t)2+(0.8−0.08t)2+(1−0.24t)2 M1A1
(=√8.64−2.688t+0.317t2)
[5 marks]
minimum when dDdt=0 (M1)
t = 3.83 A1
[2 marks]
0.511 (km) A1
[1 mark]