Let be the origin and let be an arbitrary triangle. The point is such that
·+·=·+·=·+·
Then the triangle has as its
(A)centroid
(B)circumcentre
(C)incentre
(D)orthocenterCorrect
Answer: (D)
Solution
Write for the position vectors. The first equality: $Missing argument for \vec\vec p·Missing argument for \vec\vec q+Missing argument for \vec\vec r·Missing argument for \vec\vec s-Missing argument for \vec\vec r·Missing argument for \vec\vec p-Missing argument for \vec\vec q·Missing argument for \vec\vec s=0⇒(Missing argument for \vec\vec p-Missing argument for \vec\vec
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.