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Standard XII

Mathematics

Section Formula

If A a ,B b...

Question

If $A (\to a), B (\to b), C (\to c)$ be the vertices of a triangle whose circumcentre is the origin, then orthocenter is given by

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

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

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\to a + \to b + \to c

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None of these

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Solution

The correct option is B $\to a + \to b + \to c$
$A (\to a), B (\to b), C (\to c)$ ate the vertices of the $△ A B C$ .
Circumcentre is at origin $G (\to g) = \frac{\to a + \to b + \to c}{3}$
$G = \frac{2 \to c + \to O}{3} \Rightarrow \to O = 3 \to G$
$\Rightarrow \to O = 3 ⎛ ⎝ \frac{\to a + \to b + \to c}{3} ⎞ ⎠ = \to a + \to b + \to c$

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If A(→a),B(→b),C(→c) be the vertices of a triangle whose circumcentre is the origin, then orthocenter is given by

The correct option is B →a+→b+→cA(→a),B(→b),C(→c) ate the vertices of the △ABC.Circumcentre is at origin G(→g)=→a+→b+→c3G=2→c+→O3⇒→O=3→G⇒→O=3⎛⎝→a+→b+→c3⎞⎠=→a+→b+→c

If $A (\to a), B (\to b), C (\to c)$ be the vertices of a triangle whose circumcentre is the origin, then orthocenter is given by