If G is the centroid of the triangle ABC, then $\frac{AG^2+ BG^2+CG^2}{AB^2 + BC^3+CA^2} =$
Answer & explanation
Correct answer: option 2
Taking G as the origin, let the position vectors of A, B and C be $\vec a,\vec b$ and $\vec c$ respectively. Then,
$\frac{\vec a+\vec b+\vec c}{3}=\vec 0⇒\vec a+\vec b+\vec c=\vec 0$
$⇒|\vec a+\vec b+\vec c|=0⇒|\vec a+\vec b+\vec c|^2=0$
$⇒|\vec a|^2+|\vec b|^2+|\vec c|^2=-2(\vec .\vec b+\vec b.\vec c+\vec c.\vec a)$ ...(i)
$∴AB^2 + BC^2 + CA^2$
$=|\vec{AB}|^2 +|\vec{BC}|^2 +|\vec{CA}^2|$
$=|\vec b-\vec a|^2+|\vec c-\vec b|^2+|\vec a-\vec c|^2$
$=\{|\vec a|^2+|\vec b|^2+|\vec c|^2\}-(\vec .\vec b+\vec b.\vec c+\vec c.\vec a)$
$=3\{|\vec a|^2+|\vec b|^2+|\vec c|^2\}$ [Using (i)]
$=3\{AG^2+ BG^2+CG^2\}$
Hence, $\frac{AG^2+ BG^2+CG^2}{AB^2 + BC^3+CA^2} =\frac{1}{3}$