question_answer

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)  \[2i-3j+k\]

Correct Answer: A

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 \[=2i+3j+k\]