JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 12)
Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$ respectively, then the position vector of the orthocentre of this triangle, is :
$${\overrightarrow a + \overrightarrow b + \overrightarrow c }$$
$$ - \left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$$
$$\overrightarrow 0 $$
$$\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$$
Explanation
_10th_April_Morning_Slot_en_12_1.png)
Given,
Position vector of circumcentre, $$\overrightarrow C = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$
We know, position vector of centroid, $$\overrightarrow G = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 3}$$
Now, let $$\overrightarrow R $$ be the orthocentre of the triangle.
We know, $$\overrightarrow G $$ $$ = {{2\overrightarrow C + \overrightarrow R } \over 3}$$
$$ \Rightarrow $$ 3$$\overrightarrow G $$ $$ = 2\overrightarrow C + \overrightarrow R $$
$$ \Rightarrow $$ $$\overrightarrow R = 3\overrightarrow G - 2\overrightarrow C $$
= $$\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) - 2\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}} \right)$$
= $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$$
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