It O is circumcenter and H is orthocenter, G is the centroid of ΔABC then List – I List – IIA)OA→+OB→+OC→P)12HO→B)HA→+HB→+HC→Q)2HO→C)AH→+HB→+HC→R)2AO→D)OG→S)13OH→ T)OH→The correct matching

HA→+2HD→=HA→+2(HO→+OD→) = HA→+2HO→+2OD→=2HO→ (2OD→=AH→)OG→=OA→+2OD→3⇒3OG→=OA→+2OD→ = HG→+OG→ = HO→

It O is circumcenter and H is orthocenter, G is the centroid of ΔABC then List – I List – IIA)OA→+OB→+OC→P)12HO→B)HA→+HB→+HC→Q)2HO→C)AH→+HB→+HC→R)2AO→D)OG→S)13OH→ T)OH→The correct matching

HA→+2HD→=HA→+2(HO→+OD→) = HA→+2HO→+2OD→=2HO→ (2OD→=AH→)OG→=OA→+2OD→3⇒3OG→=OA→+2OD→ = HG→+OG→ = HO→

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It O is circumcenter and H is orthocenter, G is the centroid of $Δ A B C$ then
List – I List – II
A) $\vec{O A} + \vec{O B} + \vec{O C}$ P) $\frac{1}{2} \vec{H O}$
B) $\vec{H A} + \vec{H B} + \vec{H C}$ Q) $2 \vec{H O}$
C) $\vec{A H} + \vec{H B} + \vec{H C}$ R) $2 \vec{A O}$
D) $\vec{O G}$ S) $\frac{1}{3} \vec{O H}$
T) $\vec{O H}$
The correct matching

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$A \to P, B \to T, C \to R, D \to Q$

$A \to T, B \to Q, C \to S, D \to R$

$A \to T, B \to Q, C \to R, D \to S$

$A \to Q, B \to S, C \to R, D \to T$

answer is C.

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Detailed Solution

$\begin{array}{l} \vec{H A} + 2 \vec{H D} = \vec{H A} + 2 (\vec{H O} + \vec{O D}) \\ = \vec{H A} + 2 \vec{H O} + 2 \vec{O D} = 2 \vec{H O} (2 \vec{O D} = \vec{A H}) \\ \vec{O G} = \frac{\vec{O A} + 2 \vec{O D}}{3} \Rightarrow 3 \vec{O G} = \vec{O A} + 2 \vec{O D} \\ = \vec{H G} + \vec{O G} \\ = \vec{H O} \end{array}$

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	List – I		List – II
A)	$\vec{O A} + \vec{O B} + \vec{O C}$	P)	$\frac{1}{2} \vec{H O}$
B)	$\vec{H A} + \vec{H B} + \vec{H C}$	Q)	$2 \vec{H O}$
C)	$\vec{A H} + \vec{H B} + \vec{H C}$	R)	$2 \vec{A O}$
D)	$\vec{O G}$	S)	$\frac{1}{3} \vec{O H}$
		T)	$\vec{O H}$

It O is circumcenter and H is orthocenter, G is the centroid of ΔABC then List – I List – IIA)OA→+OB→+OC→P)12HO→B)HA→+HB→+HC→Q)2HO→C)AH→+HB→+HC→R)2AO→D)OG→S)13OH→ T)OH→The correct matching

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