Q. The position vector of the centroid of the $ \Delta ABC $ is $ 2i+4j+2k $ . If the position vector of the vertex A is $ 2i+6j+4k, $ then the position vector of midpoint of BC is
KEAMKEAM 2011
Solution:
Given, the position vector of vertex A
$=2i+6j+4k $ and centroid of $ \Delta ABC $
$=2i+4j+2k. $ We know that the median AM of $ \Delta ABC $ divided by centroid G, in the ratio $ 2:1 $ .
Then, by section formula $ (2,4,2)=\left\{ \frac{2x+2}{2+1},\frac{2y+6}{2+1},\frac{2z+4}{2+1} \right\} $ On comparing,
$ \Rightarrow $ $ 2x+2=6 $
$ \Rightarrow $ $ x=2 $
$ \Rightarrow $ $ 2y+6=12 $
$ \Rightarrow $ $ y=3 $
$ \Rightarrow $ $ 22+4=6 $
$ \Rightarrow $ $ z=1 $ So, the position vector of M i.e., midpoint of BC is
$=2i+3j+k $
