Let ABC be a triangle whose circumcentre is at P, if the position vectors A,B,C and P are a⃗,b⃗,c⃗ and a→+b→+c→4 respectively. then the position vector of the orthocentre of this triangle is

A(a¯),B(b¯),C(c¯),Pa¯+b¯+c¯4 OG¯=a¯+b¯+c¯3 OP¯=a¯+b¯+c¯4 2s¯+1(0)3=a¯+b¯+c¯32S¯+O¯=a¯+b¯+c¯ O¯=a¯+b¯+c¯ =2(a¯+b¯+c¯)-(a¯+b¯+c¯) O¯=a¯+b¯+c¯

Let ABC be a triangle whose circumcentre is at P, if the position vectors A,B,C and P are a⃗,b⃗,c⃗ and a→+b→+c→4 respectively. then the position vector of the orthocentre of this triangle is

A(a¯),B(b¯),C(c¯),Pa¯+b¯+c¯4 OG¯=a¯+b¯+c¯3 OP¯=a¯+b¯+c¯4 2s¯+1(0)3=a¯+b¯+c¯32S¯+O¯=a¯+b¯+c¯ O¯=a¯+b¯+c¯ =2(a¯+b¯+c¯)-(a¯+b¯+c¯) O¯=a¯+b¯+c¯

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$Let A B C be a triangle whose circumcentre is at P, if the position vectors A,B,C and P are a⃗,b⃗,c⃗ and \frac{\vec{a} + \vec{b} + \vec{c}}{4} r e s p e c t i v e l y . t h e n t h e p o s i t i o n v e c t o r o f t h e o r t h o c e n t r e o f t h i s t r i a n g l e i s$

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$\vec{0}$

$\vec{a} + \vec{b} + \vec{c}$

$\frac{\vec{a} + \vec{b} + \vec{c}}{2}$

$\frac{\vec{a} + \vec{b} + \vec{c}}{3}$

answer is A.

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Detailed Solution

$A (\bar{a}), B (\bar{b}), C (\bar{c}), P (\frac{\bar{a} + \bar{b} + \bar{c}}{4}) \bar{O G} = \frac{\bar{a} + \bar{b} + \bar{c}}{3} \bar{O P} = \frac{\bar{a} + \bar{b} + \bar{c}}{4}$

$\frac{2 \bar{s} + 1 (0)}{3} = \frac{\bar{a} + \bar{b} + \bar{c}}{3}$

$2 \bar{S} + \bar{O} = \bar{a} + \bar{b} + \bar{c} \bar{O} = \bar{a} + \bar{b} + \bar{c} = 2 (\bar{a} + \bar{b} + \bar{c}) - (\bar{a} + \bar{b} + \bar{c}) \bar{O} = \bar{a} + \bar{b} + \bar{c}$