Vector Algebra
Scalar Product and Geometric Applications
Grade 12
Question:
<p>Let O be the origin and let PQR be an arbitrary triangle. The point S is such that \[\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS} = \overrightarrow{OQ} \cdot \overrightarrow{OR} + \overrightarrow{OP} \cdot \overrightarrow{OS}\]Then the triangle PQR has S as its</p>
<p>(a) centroid</p>
<p>(b) orthocentre</p>
<p>(c) incentre</p>
<p>(d) circumcentre</p>
Step-by-Step Solution
Key Concept: The given dot product conditions imply perpendicularity relationships that characterize the orthocentre where altitudes intersect.
From the given condition: \[\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS}\]This simplifies to: \[\overrightarrow{OS} \cdot (\overrightarrow{OR} - \overrightarrow{OQ}) = \overrightarrow{OP} \cdot (\overrightarrow{OR} - \overrightarrow{OQ})\]Therefore: \[(\overrightarrow{OS} - \overrightarrow{OP}) \cdot (\overrightarrow{OR} - \overrightarrow{OQ}) = 0\]This means \(\overrightarrow{PS} \perp \overrightarrow{QR}\). Similarly, from the other equalities, we can show that S is equidistant (in the perpendicular sense) from all sides, making it the orthocentre .
Correct Answer: B