Let ABC be a triangle whose circumcentre is at P. If the position vectors of A,B,C and P are a,b,c and a+b+c4 respectively, then the position vector of the ortho-centre of this triangle, is

OA¯=a¯,OB¯=b¯,OC¯=c¯,OP¯=a¯+b¯+c¯4OG¯=a¯+b¯+c¯3If H is orthocenter then 3G¯=2P¯+H¯ where P is circumcentre⇒H¯=(a¯+b¯+c¯)−2(a¯+b¯+c¯4)=a¯+b¯+c¯2

Let ABC be a triangle whose circumcentre is at P. If the position vectors of A,B,C and P are a,b,c and a+b+c4 respectively, then the position vector of the ortho-centre of this triangle, is

OA¯=a¯,OB¯=b¯,OC¯=c¯,OP¯=a¯+b¯+c¯4OG¯=a¯+b¯+c¯3If H is orthocenter then 3G¯=2P¯+H¯ where P is circumcentre⇒H¯=(a¯+b¯+c¯)−2(a¯+b¯+c¯4)=a¯+b¯+c¯2

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Let ABC be a triangle whose circumcentre is at P. If the position vectors of A,B,C and P are $a, b, c$ and $\frac{a + b + c}{4}$ respectively, then the position vector of the ortho-centre of this triangle, is

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

$a + b + c$

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

$O$

answer is C.

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Detailed Solution

$\begin{array}{l} \bar{OA} = \bar{a}, \bar{OB} = \bar{b}, \bar{OC} = \bar{c}, \bar{OP} = \frac{\bar{a} + \bar{b} + \bar{c}}{4} \\ \bar{OG} = \frac{\bar{a} + \bar{b} + \bar{c}}{3} \end{array}$
If H is orthocenter then $\bar{3 G} = 2 \bar{P} + \bar{H}$ where P is circumcentre
$\Rightarrow \bar{H} = (\bar{a} + \bar{b} + \bar{c}) - 2 (\frac{\bar{a} + \bar{b} + \bar{c}}{4}) = \frac{\bar{a} + \bar{b} + \bar{c}}{2}$