(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) Dete...

Answer Details

(a)(i) With \( A(1,5),\ B(3,9),\ C(-1,1) \):

\( \vec{AB} = B-A = (2,\ 4) \) and \( \vec{BC} = C-B = (-4,\ -8) \).

Since \( \vec{BC} = -2\,\vec{AB} \), the two vectors are parallel and share the common point B. Therefore A, B and C are collinear.

(ii) \( |AB| = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt5 \) and \( |BC| = \sqrt{(-4)^2+(-8)^2} = \sqrt{80} = 4\sqrt5 \).

\[ |AB| : |BC| = 2\sqrt5 : 4\sqrt5 = 1 : 2. \]

(b) Let the beam XY, length 24 m, lie horizontally. String tensions: \(T_X\) at X and \(T_M\) at M, where M is 4 m from Y, i.e. 20 m from X. The beam weight \(10g\) acts at the centre, 12 m from X; the 50 kg load acts at N, 8 m from X (take \(g = 10\text{ ms}^{-2}\)).

Vertical equilibrium: \( T_X + T_M = (10+50)g = 60g \).

Moments about X: \[ T_M(20) = 10g(12) + 50g(8) = 120g + 400g = 520g \Rightarrow T_M = 26g = 260\text{ N}. \]

Then \( T_X = 60g - 26g = 34g = 340\text{ N} \).

The tension at X is \(340\text{ N}\) and the tension at M is \(260\text{ N}\).