(a) The position vectors of points A, B and C are \(i + 5j , 3i + 9j\) and \(-i + j\) respectively. (i) Show that points A, B and C are collinear; (ii) Determine the ratio \(|AB| : |BC|\).
(b) A uniform beam XY of mass 10 kg and length 24m is hunged horizontally from a cross bar by teo vertical inextensible strings, one attached to X and the other at a point M, 4m away from Y. A mass of 50kg is suspended at a point N which is 8m from X. If the system remains in equilibrium, calculate the tensions in the strings.
(a)(i) Position vector of A, \(\overrightarrow{OA} = i + 5j\)
Position vector of B, \(\overrightarrow{OB} = 3i + 9j\)
Position vector of C, \(\overrightarrow{OC} = -i + j\)
The points A, B and C are collinear if \(|\overrightarrow{AB}| = k|\overrightarrow{BC}|\), where k is a constant.
If k > 0, then \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are in the same direction.
If k < 0, then they are in opposite directions, in this case they are A, C, B.
\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = - \overrightarrow{OA} + \overrightarrow{OB}\)
= \(- i - 5j + (3i + 9j) = 2i + 4j = 2(i + 2j)\)
\(\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = - \overrightarrow{OB} + \overrightarrow{OC}\)
= \(- 3i - 9j + (-i + j) = - 4i - 8j = -4(i + 2j)\)
\(\overrightarrow{AB} = -\frac{1}{2} \overrightarrow{BC}\)
\(\therefore\) A, B and C are collinear with \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) in opposite directions.
(ii) \(|\overrightarrow{AB}| : |\overrightarrow{BC}| = 1 : 2\)
(b) 
We take moments about X.
Clockwise moment = \(500 \times 8 + 100 \times 12 = 5200N\)
Anticlockwise moment = \(T_{2} \times 20\)
From the principles of moment,
\(20T_{2} = 5200N \implies T_{2} = 260N\)
For vertical equilibrium,
\(T_{1} + T_{2} = 600N\)
\(\therefore T_{1} = 600 - T_{2} = 600 - 260 =340N\)
The tensions in the strings are 340N and 260N respectively.
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