A drone is flying over Bengaluru, monitoring traffic. Its position relative to a control tower is given by r(t)=(3t)i^+(4t)j^+(50−5t)k^ where t is in seconds and distances in metres. (a) Find the drone’s velocity vector. (b) Find its speed. (c) Find the time at which the drone is closest to the tower. (d) Find the minimum distance.
Solution — Step by Step
v(t)=dtdr=3i^+4j^−5k^
The velocity is constant — drone moves with uniform velocity.
Minimum distance: 50⋅25−500⋅5+2500=1250−2500+2500=1250=252 m.
(a) v=3i^+4j^−5k^, (b) speed =52 m/s, (c) at t=5 s, (d) minimum distance = 252 m.
Why This Works
For a particle moving in a straight line with constant velocity, the closest approach to a fixed point happens when the position vector is perpendicular to the velocity vector. Equivalently, dtd∣r∣2=0.
You can verify the perpendicularity: r(5)=(15,20,25) and v=(3,4,−5). Dot product: 15⋅3+20⋅4+25⋅(−5)=45+80−125=0. Confirmed.
For “minimum distance” or “closest approach” problems, always set dtd∣r∣2=0 instead of dtd∣r∣=0. Squared distance avoids the square root and makes differentiation cleaner.
Alternative Method
Geometric approach: foot of perpendicular from the origin onto the line. The line has direction v=(3,4,−5) and passes through (0,0,50) at t=0. The closest point: r0+t(v) where t=−r0⋅v/∣v∣2=−(0,0,50)⋅(3,4,−5)/50=250/50=5. Same answer: t=5 s.
Common Mistake
Students compute ∣r∣ first and try to differentiate the square root. Always work with ∣r∣2 for minimization — the same t minimizes both, and the algebra is cleaner.
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