The position vectors of two vertices A and B of a triangle ABC are \(\hat{\imath}+2\hat{\jmath}-3\hat{k}\) and \(3\hat{\imath}-\hat{\jmath}+2\hat{k}\) respectively. If the centroid of the triangle has position vector \(2\hat{\imath}+\hat{\jmath}\), then the position vector of the vertex C is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2\hat{\imath}+2\hat{\jmath}+\hat{k}\)
The centroid G of a triangle with vertices of position vectors \(\vec{a}, \vec{b}, \vec{c}\) has position vector
\( \displaystyle \vec{g} = \frac{\vec{a}+\vec{b}+\vec{c}}{3} \)
So
\(\vec{c} = 3\vec{g}-\vec{a}-\vec{b}\)
\(3\vec{g} = 6\hat{\imath}+3\hat{\jmath}\)
\(\vec{a}+\vec{b} = 4\hat{\imath}+\hat{\jmath}-\hat{k}\)
\(\vec{c} = (6-4)\hat{\imath}+(3-1)\hat{\jmath}+(0+1)\hat{k}\)
\( = 2\hat{\imath}+2\hat{\jmath}+\hat{k}\)