Vector Algebra
Vector Relations in Triangles
Grade 12
Question:
<p>If O be the circumcentre and O' be the orthocentre of the $\triangle ABC$, then $\vec{O'A} + \vec{O'B} + \vec{O'C}$ is equal to</p>
<p>(a) $\vec{OO'}$</p>
<p>(b) $2\vec{O'O}$</p>
<p>(c) $2\vec{OO'}$</p>
<p>(d) $0$</p>
Step-by-Step Solution
Key Concept: Use vector addition from O' to vertices via circumcentre O. The key property is that for circumcentre O, $\vec{OA} + \vec{OB} + \vec{OC} = \vec{OO'}$ where O' is the orthocentre.
Solution: We have: $\vec{O'A} = \vec{O'O} + \vec{OA}$ $\vec{O'B} = \vec{O'O} + \vec{OB}$ $\vec{O'C} = \vec{O'O} + \vec{OC}$ Adding these three equations: $\vec{O'A} + \vec{O'B} + \vec{O'C} = 3\vec{O'O} + (\vec{OA} + \vec{OB} + \vec{OC})$ Since O is the circumcentre, $\vec{OA} + \vec{OB} + \vec{OC} = \vec{OO'} = -\vec{O'O}$ Therefore: $\vec{O'A} + \vec{O'B} + \vec{O'C} = 3\vec{O'O} - \vec{O'O} = 2\vec{O'O}$
Correct Answer: b