Vectors in two and three dimensions — Question 249
Original QCE Vault practice · 12 marks
Q249 · Practice questionTechnology-activeVery complex unfamiliar12 marks
QUESTION 249 (12 marks)
A flexible cable runs from A=(1,1,4) to B=(5,−1,2) through a junction P on the floor z=0. Coordinates are in metres. Its length is ∣AP∣+∣PB∣. The junction is first free to move anywhere on the floor, then is constrained to the rail x+y=2, z=0.
a)
For the unrestricted floor junction, determine the shortest cable length and the position of P. Justify that your solution is a global minimum.
[4 marks]
b)
For the rail constraint, let P=(s,2−s,0). Show that the length is F(s)=2(s−1)2+16+2(s−4)2+6.
[2 marks]
c)
Determine the rail junction that minimises the length. Give an exact value of s and the minimum length to three decimal places. Prove that the minimum is unique.
(a) P=(11/3,−1/3,0) and minimum length 214 m. (b) The stated expression for F(s). (c) s=4+616+6; minimum 7.720 m; unique.
Worked solution
(a) Reflect A in the floor to A∗=(1,1,−4). For every floor point P, ∣AP∣=∣A∗P∣, so the triangle inequality gives ∣AP∣+∣PB∣≥∣A∗B∣=42+(−2)2+62=214. Equality occurs on the segment A∗+t(4,−2,6), which meets z=0 at t=2/3. This yields P=(11/3,−1/3,0) and attains the lower bound.
(b) The squared distances are (s−1)2+(1−s)2+16=2(s−1)2+16 and (s−5)2+(3−s)2+4=2(s−4)2+6. Taking square roots and adding gives F.
(c) F′(s)=2(s−1)/2(s−1)2+16+2(s−4)/2(s−4)2+6. It is negative for s≤1 and positive for s≥4, so a stationary point lies between 1 and 4. On that interval, F′=0 is equivalent to 6(s−1)=4(4−s) after squaring and restoring the positive signs. Thus s=(16+6)/(4+6). Moreover, F′′=32/[2(s−1)2+16]3/2+12/[2(s−4)2+6]3/2>0 for every real s. The stationary point is therefore the unique global minimum. Substitution gives F=40+86=7.719839… m.
