Single Choice

If $$A(\overline{a})$$ , $$B(\overline{b})$$ and $$C(\overline{c})$$ be the vertices of a triangle $$ABC$$ whose circumcentre is the origin then orthocentre is given by

A$$\overline{ a}+\overline{b}+\overline{c}$$
Correct Answer
B$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{3}$$
C$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{2}$$
D$$\displaystyle \dfrac{\overline{a}+\overline{b}+\overline{c}}{4}$$

Solution

P.V. of circumcentre $$= (0,0,0)$$
P.V. of circumcentre $$= \dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$$
$$centroid = \dfrac{orthocentre + 2\ circumcentre}{3}$$ (section formula)
$$\dfrac{\vec{a}+\vec{b}+\vec{c}}{3}= \dfrac{orthocentre}{3}$$
orthocentre$$ = \vec{a}+\vec{b}+\vec{c}$$


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