Vector Algebra
Vector Algebra
nta_pyq_2025_jan
Grade 12
Question:
Let the position vectors of three vertices of a triangle be 4p\to + q\to - 3r\to, -5p\to + q\to + 2r\to and 2 p - q + 2 r . If the \to \to \to position vectors of the orthocenter and the circumcenter of the triangle are and \alphap\to + \betaq\to + \gammar\to p +q +r 4 respectively, then \alpha + 2\beta + 5\gamma is equal to :
Step-by-Step Solution
Key Concept: Apply the core result for dot product, cross product and projections and simplify using the given constraints.
We know that (1) \to \to \to p + q + r O ( (orthocentre) 4 \to \to \to C( circum centre) \alpha p + \beta q + \gamma r \to \to \to p + q + r C (centroid) = 3 by relation \to \to \to \to \to \to p + q + r p + q + r \to \to \to \Rightarrow 2(\alpha p + \beta q + \gamma r ) + = 3( ) 4 3 \to \to \to \to \to \to \Rightarrow 8(\alpha p + \beta q + \gamma r ) = 3( p + q + r ) \Rightarrow 8\alpha = 3, 8\beta = 3, 8\gamma = 3 3 3 3 \alpha = ,\beta = ,\gamma = 8 8 8 \therefore \alpha + 2\beta + 3\gamma 3 6 15 24 + + = = 3 8 8 8 8 \to = ^i + ^j + k
Correct Answer: 1