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Collinear Vectors

Let ABC be a ...

Question

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

A

\to a + \to b + \to c

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B

\to 0

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C

- (\frac{\to a + \to b + \to c}{2})

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D

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

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Solution

The correct option is D $\frac{\to a + \to b + \to c}{2}$
Position vector of centroid of $△ A B C = \frac{\to a + \to b + \to c}{3}$
We know that orthocentre, centroid and circumcentre of any triangle are collinear and the centroid divides the line-segment joining orthocentre and circumcentre in the ratio $2 : 1$ .
$∴ \frac{\to a + \to b + \to c}{3} = \frac{2 (\frac{\to a + \to b + \to c}{4}) + 1 (P.V. of orthocentre)}{2 + 1}$
$\Rightarrow P.V. of orthocentre = \frac{\to a + \to b + \to c}{2}$

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Collinear Vectors

Standard XII Mathematics

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