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Question: Let the unit vectors \(\overrightarrow{a}\), \(\overrightarrow{b}\), \(\overrightarrow{c}\) be the p...

Let the unit vectors a→\overrightarrow{a}, b→\overrightarrow{b}, c→\overrightarrow{c} be the position vectors of the vertices of a triangle ABC. If F→\overrightarrow{F} is the position vector of the mid point of the line segment joining its orthocentre and centroid then (a→\overrightarrow{a}–F→\overrightarrow{F})2 + (b→\overrightarrow{b}–F→\overrightarrow{F})2 + (c→\overrightarrow{c}–F→\overrightarrow{F})2 =

A

1

B

2

C

3

D

None of these

Answer

3

Explanation

Solution

Let the circumcentre of the triangle be the origin.

Ž orthocentre is a→\overrightarrow{a}+ b→\overrightarrow{b}+ c→\overrightarrow{c}and the centroid is a→+b→+c→3\frac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3}

Ž F = 23\frac{2}{3} (a→\overrightarrow{a}+ b→\overrightarrow{b}+ c→\overrightarrow{c})

Ž (a→\overrightarrow{a}–F→\overrightarrow{F})2 + (b→\overrightarrow{b} –F→\overrightarrow{F})2 + (c→\overrightarrow{c}–F→\overrightarrow{F})2

=19\frac{1}{9} [(a→\overrightarrow{a}– 2(b→\overrightarrow{b}+c→\overrightarrow{c})2 ) + (b→\overrightarrow{b}– 2(a→\overrightarrow{a}+c→\overrightarrow{c})2) + (c→\overrightarrow{c}–2 (a→\overrightarrow{a}+b→\overrightarrow{b})2)]

= 19\frac{1}{9} (27) = 3