If G( vecg), H( vech) and P( vecp) are centroid, orthocenter and circumcenter of a triangle and x vecp+y vech+z vecg = 0 then (x, y, z) =

Question

If G( vecg), H( vech) and P( vecp) are centroid, orthocenter and circumcenter of a triangle and x vecp+y vech+z vecg = 0 then (x, y, z) =  (A) 1, 1, - 2  (B) 2, 1, -3 (C) 1, 3, -4 (D) 2, 3, -5

Tardigrade Admin · Accepted Answer

We know that, orthocentre, centroid and circumcentre of a triangle are collinear and centroid divides orthocentre and circumcentr in the ratio 2: 1
By using internally division,

\frac{2 p + 1 h}{2 + 1} = g

\Rightarrow 2 p + h - 3 g = 0

But it is given

x p + y h + z g = 0

x = 2, y = 1

and

z = - 3

Q. If $G (g), H (h)$ and $P (p)$ are centroid, orthocenter and circumcenter of a triangle and $x p + y h + z g = 0$ then $(x, y, z) =$ _____________

$1, 1, - 2$

$2, 1, - 3$

$1, 3, - 4$

$2, 3, - 5$

Q. If G(g​),H(h) and P(p​) are centroid, orthocenter and circumcenter of a triangle and xp​+yh+zg​=0 then (x,y,z)= _____________

1,1,−2

2,1,−3

1,3,−4

2,3,−5

We know that, orthocentre, centroid and circumcentre of a triangle are collinear and centroid divides orthocentre and circumcentr in the ratio 2: 1 By using internally division, 2+12p​+1h​=g ⇒2p​+h−3g​=0 But it is given xp​+yh+zg​=0 x=2,y=1 and z=−3

Q. If $G (g), H (h)$ and $P (p)$ are centroid, orthocenter and circumcenter of a triangle and $x p + y h + z g = 0$ then $(x, y, z) =$ _____________

$1, 1, - 2$

$2, 1, - 3$

$1, 3, - 4$

$2, 3, - 5$