(a) Two ships M and N, moving with constant velocities, have position vectors (3i +
(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, = \(6i + 9j\)
\(\overrightarrow{MN} = (6i + 9j) - (15i + 18j)\)
= \(-9i - 9j\)
\(|MN| = \sqrt{(-9)^{2} + (-9)^{2}} = \sqrt{162}\)
= \(9\sqrt{2} m\)
(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,
\(i : 5 + q - 2p - 4 = 0 \implies q - 2p = -1 ...(1)\)
\(j : p + 1 + 3 + q = 0 \implies p + q = -4 ... (2)\)
\((2) - (1) : 3p = -3 \implies p = -1\)
\(-1 + q = -4 \implies q = -3\)