Let origin O be the orthocentre of an equilateral triangle ABC- If OA - a- OB - b- OC - c - then AB -2BC-3CA-

AB +2BC+3CA = b a+2(c b) + 3(a c) = b + 2a c⋯ (i) For an equilateral centriod is equal to orthocentre, 0 = a+b+c3 b = a+c Equation (i) becomes AB +2BC+3CA ...

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Standard XII

Mathematics

Distances between Special Points in a Triangle

Let origin O ...

Question

Let origin $O$ be the orthocentre of an equilateral triangle $A B C$ . If $- - \to O A = \to a, - - \to O B = \to b, - - \to O C = \to c,$ then $- - \to A B + 2 - - \to B C + 3 - - \to C A =$

3 \to c

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3 \to a

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\to 0

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3 \to b

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Solution

The correct option is B $3 \to a$
$- - \to A B + 2 - - \to B C + 3 - - \to C A$
$= \to b - \to a + 2 (\to c - \to b) + 3 (\to a - \to c)$
$= - \to b + 2 \to a - \to c \dots (i)$
For an equilateral centriod is equal to orthocentre,
$0 = \frac{\to a + \to b + \to c}{3}$
$- \to b = \to a + \to c$
Equation $(i)$ becomes
$- - \to A B + 2 - - \to B C + 3 - - \to C A = \to a + \to c + 2 \to a - \to c$
$- - \to A B + 2 - - \to B C + 3 - - \to C A = 3 \to a$

Suggest Corrections

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Distances between Special Points in a Triangle

Standard XII Mathematics

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Let origin O be the orthocentre of an equilateral triangle ABC. If −−→OA=→a,−−→OB=→b,−−→OC=→c, then −−→AB+2−−→BC+3−−→CA=

Let origin $O$ be the orthocentre of an equilateral triangle $A B C$ . If $- - \to O A = \to a, - - \to O B = \to b, - - \to O C = \to c,$ then $- - \to A B + 2 - - \to B C + 3 - - \to C A =$