Vector Algebra
Triangle Properties
Grade 12
Question:
<p><strong>Ex. 62:</strong> Let ABC be a triangle whose centroid is G, orthocentre is H and circumcentre is the origin 'O'. If D is any point in the plane of the triangle such that no three of O, A, C and D are collinear satisfying the relation \(\vec{AD} + \vec{BD} + \vec{CH} + 3\vec{HG} = \lambda \vec{HD}\), then what is the value of the scalar \(\lambda\)?</p>
Step-by-Step Solution
Key Concept: Use the properties of centroid and position vectors to simplify the given vector equation.
Step 1: Expand the left-hand side using position vectors. \[\text{LHS} = (\vec{d} - \vec{a}) + (\vec{d} - \vec{b}) + (\vec{h} - \vec{c}) + 3(\vec{g} - \vec{h})\] Step 2: Use the fact that centroid \(\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}\). \[= 2\vec{d} - (\vec{a} + \vec{b} + \vec{c}) + 3 \cdot \frac{\vec{a} + \vec{b} + \vec{c}}{3} - 2\vec{h}\] \[= 2\vec{d} - 2\vec{h} = 2(\vec{d} - \vec{h}) = 2\vec{HD}\] Step 3: Compare with \(\lambda \vec{HD}\) to get \(\lambda = 2\).
Correct Answer: 2