Q. Let the position vectors of three vertices of a triangle be $4\vec p+\vec q-3\vec r$, $-5\vec p+\vec q+2\vec r$ and $2\vec p-\vec q+2\vec r$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec p+\vec q+\vec r}4$ and $\alpha\vec p+\beta\vec q+\gamma\vec r$ respectively, then $\alpha+2\beta+5\gamma$ is equal to :
A) 3
B) 4
C) 1
D) 6
Explanation
[Solution by Dr. Abubacker] The sum of the vertices is $\vec p+\vec q+\vec r$, so the centroid is $G=\frac{\vec p+\vec q+\vec r}3$. Since $H=3G-2O$: $O=\frac{3G-H}2=\frac12\left(1-\frac14\right)(\vec p+\vec q+\vec r)=\frac38(\vec p+\vec q+\vec r)$. So $\alpha=\beta=\gamma=\frac38$ and $\alpha+2\beta+5\gamma=8\cdot\frac38=3$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.