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KEAM 2011 Math Paper

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Question : 23 of 120

Marks: +1, -0

The position vector of the centroid of the triangle ABC is

2\hat{i}+4\hat{j}+2\hat{k}.

If the position vector of the vertex A is

2\hat{i}-6\hat{j}+4\hat{k},

9 then the position vector of midpoint of BC is

$2\hat{i}+3\hat{j}+\hat{k}$
$2\hat{i}+3\hat{j}-\hat{k}$
$2\hat{i}-3\hat{j}-\hat{k}$
$-2\hat{i}-3\hat{j}-\hat{k}$
$2\hat{i}-3\hat{j}+\hat{k}$

Solution:

Given, the position vector of vertex

A

=2i+6j+4k

and centroid of

\triangle ABC

=2i+4j+2k

We know that the median

A M

of

\triangle 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 \;\; 2z+4=6

\Rightarrow \;\; z=1

So, the position vector of

M

i.e., mid point of

B C

is

=2i+3j+k

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