<p>Three forces <span>\(\vec{P}\)</span>, <span>\(\vec{Q}\)</span> and <span>\(\vec{R}\)</span> acting along <span>\(IA\)</span>, <span>\(IB\)</span> and <span>\(IC\)</span>, where <span>\(I\)</span> is the incentre of a <span>\(\triangle ABC\)</span>, are in equilibrium. Then <span>\(\vec{P} : \vec{Q} : \vec{R}\)</span> is</p>
<p>\(\cos\dfrac{A}{2} : \cos\dfrac{B}{2} : \cos\dfrac{C}{2}\)</p>
<p>\(\sin\dfrac{A}{2} : \sin\dfrac{B}{2} : \sin\dfrac{C}{2}\)</p>
<p>Option 3</p>
<p>Option 4</p>
Step-by-Step Solution
Key Concept: For forces in equilibrium along directions from incentre I to vertices, use the property that the incentre divides the angle bisectors in a specific ratio related to side lengths. The magnitudes of equilibrium forces along IA, IB, IC are proportional to the sines of half-angles at those vertices, which equals the ratio of opposite sides.
Step 1: Since I is the incentre of △ABC, the directions IA, IB, and IC correspond to angle bisectors from I to vertices. Step 2: For three forces in equilibrium acting from point I along IA, IB, IC, we use: P⃗ + Q⃗ + R⃗ = 0⃗ Step 3: The incentre I divides the triangle such that forces along these directions balance when their magnitudes are proportional to the sines of the angles they make with the equilibrium plane, which are the half-angles of the triangle. Step 4: Using the equilibrium condition and properties of the incentre: P : Q : R = sin(A/2) : sin(B/2) : sin(C/2) Step 5: This can also be expressed as: P : Q : R = 1/a : 1/b : 1/c or equivalently bc : ca : ab , depending on the convention of the answer choices. Step 6: Alternatively, using the standard result for forces in equilibrium along incentre directions: P : Q : R = a : b : c (proportional to sides opposite to vertices A, B, C respectively) ∴ Answer: A
Correct Answer: A