Question

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

• A. $2i+3j+k$
• B. $2i+3j-k$
• C. $2i-3j-k$
• D. $-2i-3j-k$
• E. $2i-3j+k$

Answer 1

Answer: (A)

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
$=2i+3j+k$