(a) Two ships M and N, moving with constant velocities, have position vectors (3i + 7j) and (4i + 5j) respectively. If the velocities of M and N are (5i + 6j) and (2i + 3j) and the distance covered by the ships after t seconds are in metres, find (i) MN ; (ii) |MN|, when t = 3 seconds.
(b) A particle is acted upon by forces (F_{1} = 5i + pj ; F_{2} = qi + j ; F_{3} = -2pi + 3j) and (F_{4} = -4i + qj), where p and q are constants. If the particle remains in equilibrium under the action of these forces, find the values of p and q.
Explanation
(a)(i) Position vector of M = 3i + 7j i.e (overrightarrow{OM} = 3i + 7j).
Position vector of N = 4i + 5j i.e (overrightarrow{ON} = 4i + 5j)
(overrightarrow{MN} = overrightarrow{ON} – overrightarrow{OM})
((4i + 5j) – (3i + 7j) = i – 2j)
Velocity of M = 5i + 6j
Distance covered in time t = ((5i + 6j)t = 5ti + 6tj)
When t = 3 seconds, = (15i + 18j)
Velocity of N = 2i + 3j
Distance covered in time t = ((2i + 3j)t = 2ti + 3tj)
When t = 3 seconds, = (6i + 9j)
(overrightarrow{MN} = (6i + 9j) – (15i + 18j))
= (-9i – 9j)
(|MN| = sqrt{(-9)^{2} + (-9)^{2}} = sqrt{162})
= (9sqrt{2} m)
(b) (F_{1} = 5i + pj ; F_{2} = qi + j ; F_{3} = -2pi + 3j ; F_{4} = -4i + qj)
If the particle remains in equilibrium, then
(F_{1} + F_{2} + F_{3} + F_{4} = 0)
(5i + pj + qi + j – 2pi + 3j – 4i + qj = 0)
Equating by components,
(i : 5 + q – 2p – 4 = 0 implies q – 2p = -1 …(1))
(j : p + 1 + 3 + q = 0 implies p + q = -4 … (2))
((2) – (1) : 3p = -3 implies p = -1)
(-1 + q = -4 implies q = -3)