Question

If \( 7i - 4j - 5k \) is the position vector of vertex A of a tetrahedron ABCD and \( -i + 4j - 3k \) is the position vector of the centroid of the triangle BCD, then the position vector of the centroid of the tetrahedron ABCD is:

• A. \( -i + 4j - 3k \)
• B. \( \frac{1}{2} (i + 4j - 3k) \)
• C. \( i + 2j + k \)
• D. \( -i + 2j + k \)

Hint:

When dealing with centroids in a tetrahedron, remember that the centroid of the tetrahedron is the average of the centroids of the triangles formed by its faces.

Answer 1

The Correct Option is C

We are given: - Position vector of vertex A: \( \vec{A} = 7i - 4j - 5k \) - Position vector of the centroid of triangle BCD: \( \vec{G}_{BCD} = -i + 4j - 3k \) The centroid \( G \) of a tetrahedron is the average of the position vectors of its vertices. The centroid of the tetrahedron ABCD is given by the average of the position vectors of \( A \), \( B \), \( C \), and \( D \). Thus, the centroid of tetrahedron ABCD is: \[ \vec{G}_{ABCD} = \frac{1}{4} (\vec{A} + \vec{B} + \vec{C} + \vec{D}) \] Since the centroid of triangle BCD is given by \( \vec{G}_{BCD} = \frac{1}{3} (\vec{B} + \vec{C} + \vec{D}) \), we can use the relationship between the centroids to compute: \[ \vec{G}_{ABCD} = \frac{1}{4} (\vec{A} + 3 \vec{G}_{BCD}) \] Substitute the given values: \[ \vec{G}_{ABCD} = \frac{1}{4} \left( (7i - 4j - 5k) + 3(-i + 4j - 3k) \right) \] Simplify: \[ \vec{G}_{ABCD} = \frac{1}{4} \left( 7i - 4j - 5k - 3i + 12j - 9k \right) \] \[ \vec{G}_{ABCD} = \frac{1}{4} \left( 4i + 8j - 14k \right) \] \[ \vec{G}_{ABCD} = i + 2j + k \] % Final Answer \[ \boxed{i + 2j + k} \]