Question

ABCD is a tetrahedron. \( \vec{i}-2\vec{j}+3\vec{k} \), \( -2\vec{i}+\vec{j}+3\vec{k} \), \( 3\vec{i}+2\vec{j}-\vec{k} \) are the position vectors of the points A, B, C respectively. \( -\vec{i}+2\vec{j}-3\vec{k} \) is the position vector of the centroid of the triangular face BCD. If G is the centroid of the tetrahedron, then GD =

• A. \( \frac{\sqrt{13}}{\sqrt{2}} \)
• B. \( \sqrt{23} \)
• C. \( \frac{\sqrt{213}}{\sqrt{2}} \)
• D. \( \sqrt{46} \)

Hint:

The centroid of a set of points is simply the average of their position vectors. For a triangle with vertices A, B, C, the centroid is \((\vec{a}+\vec{b}+\vec{c})/3\). For a tetrahedron with vertices A, B, C, D, the centroid is \((\vec{a}+\vec{b}+\vec{c}+\vec{d})/4\).

Answer 1

The Correct Option is C