If G(vec{g}), H(vec{h}) and P(vec{p}) are the | vedclass

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(B) We know that in any triangle,the centroid $(G)$,orthocenter $(H)$,and circumcenter $(P)$ are collinear,forming the Euler line.$G$ divides the line

Vedclass · Accepted Answer

$(2, 1, -3)$

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(B) We know that in any triangle,the centroid $(G)$,orthocenter $(H)$,and circumcenter $(P)$ are collinear,forming the Euler line.$G$ divides the line segment $PH$ internally in the ratio $1:2$.Using the section formula for the position vector $\vec{g}$ of the centroid $G$:$\vec{g} = \frac{1 \cdot \vec{h} + 2 \cdot \vec{p}}{1 + 2}$$\vec{g} = \frac{\vec{h} + 2\vec{p}}{3}$Multiplying both sides by $3$:$3\vec{g} = \vec{h} + 2\vec{p}$Rearranging the terms to the form $x\vec{p} + y\vec{h} + z\vec{g} = 0$:$2\vec{p} + 1\vec{h} - 3\vec{g} = 0$Comparing this with $x\vec{p} + y\vec{h} + z\vec{g} = 0$,we get $x = 2$,$y = 1$,and $z = -3$.Thus,$(x, y, z) = (2, 1, -3)$.

If $G(\vec{g}), H(\vec{h})$ and $P(\vec{p})$ are the centroid,orthocenter,and circumcenter of a triangle respectively,and $x \vec{p} + y \vec{h} + z \vec{g} = 0$,then $(x, y, z) = $

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