Question

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

• A. $1, 1, -2$
• B. $1, 3, -4$
• C. $2, 1, -3$
• D. $2, 3, -5$

Hint:

Remember the alphabetical mnemonic ordered sequence for the Euler line: H $\rightarrowG\rightarrow$ O (or P for circumcenter), with the ratio 2 : 1. Keeping the word "HG Out" or "HGP" in mind helps you write down the section formula correctly every single time!

Answer 1

The Correct Option is C

Step 1: Recall the Euler-line fact.
In any triangle the orthocenter H, centroid G, and circumcenter P are collinear, and G divides the segment from H to P in the ratio $2:1$ (with G nearer P).
Step 2: Apply the section formula.
G divides HP internally in ratio $2:1$ measured from H, so $\vec g=\dfrac{2\vec p+1\vec h}{2+1}=\dfrac{2\vec p+\vec h}{3}$.
Step 3: Clear the fraction.
Multiply by 3: $3\vec g=2\vec p+\vec h$.
Step 4: Bring all terms to one side.
$2\vec p+\vec h-3\vec g=\vec0$.
Step 5: Compare with the given form.
The given relation is $x\vec p+y\vec h+z\vec g=\vec0$. Matching coefficients term by term gives $x=2,\ y=1,\ z=-3$.
Step 6: State the answer.
So $(x,y,z)=(2,1,-3)$, matching option (3).
\[ \boxed{(x,y,z)=(2,\,1,\,-3)} \]