Question

Let the position vectors of the vertices A, B, and C of a tetrahedron ABCD be \( \hat{i} + 2\hat{j} + \hat{k} \), \( \hat{i} + 3\hat{j} - 2\hat{k} \), and \( 2\hat{i} + \hat{j} - \hat{k} \) respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is \( \frac{\sqrt{10}}{3} \) and the volume of the tetrahedron is \( \frac{\sqrt{805}}{6\sqrt{2}} \), then the position vector of E is:

• A. \( \frac{1}{2} (\hat{i} + 4\hat{j} + 7\hat{k}) \)
• B. \( \frac{1}{12} (7\hat{i} + 4\hat{j} + 3\hat{k}) \)
• C. \( \frac{1}{6} (12\hat{i} + 12\hat{j} + \hat{k}) \)
• D. \( \frac{1}{6} (7\hat{i} + 12\hat{j} + \hat{k}) \)

Hint:

To solve for position vectors in 3D geometry problems, use the properties of medians, altitudes, and perpendicularity in combination with vector operations such as dot products and cross products.

Answer 1

The Correct Option is D

We are given: - \( A(1, 2, 1) \), - \( B(1, 3, -2) \), - \( C(2, 1, -1) \), - The point \( E \) lies on the median of triangle \( ABC \), and the altitude from \( D \) intersects this line at point \( E \). 

Step 1: Calculate the area of triangle ABC. The area of \( \triangle ABC \) is given by: \[ \text{Area of } \triangle ABC = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|. \] Using the position vectors of \( A \), \( B \), and \( C \), we calculate: \[ \overrightarrow{AB} = \langle 0, 1, -3 \rangle, \quad \overrightarrow{AC} = \langle 1, -1, -2 \rangle. \] The cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \) is: \[ \overrightarrow{AB} \times \overrightarrow{AC} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 
0 & 1 & -3 
1 & -1 & -2 \end{array} \right| = \hat{i} \left( 1 \times (-2) - (-1 \times -3) \right) - \hat{j} \left( 0 \times (-2) - (1 \times -3) \right) + \hat{k} \left( 0 \times (-1) - (1 \times 1) \right) = \langle -1, 3, -1 \rangle. \] So the area is: \[ \text{Area of } \triangle ABC = \frac{1}{2} \times \sqrt{(-1)^2 + 3^2 + (-1)^2} = \frac{1}{2} \times \sqrt{11} = \frac{\sqrt{35}}{2}. \] 

Step 2: Using the volume formula. The volume \( V \) of the tetrahedron is given by: \[ V = \frac{1}{3} \times \text{Base Area} \times h. \] We are given the volume as \( \frac{\sqrt{805}}{6\sqrt{2}} \), and we already know the base area, so we solve for \( h \): \[ \frac{1}{3} \times \frac{\sqrt{35}}{2} \times h = \frac{\sqrt{805}}{6\sqrt{2}} \quad \Rightarrow \quad h = \frac{\sqrt{23}}{2}. \] 

Step 3: Calculating \( AE \). Since \( AE^2 = AD^2 - DE^2 \), we can calculate \( AE \) as: \[ AE^2 = \frac{13}{18}, \quad AE = \frac{\sqrt{13}}{18}. \] 

Step 4: Finding the position vector of E. Finally, we compute the position vector of point \( E \) as: \[ AE = \left| \mathbf{A} - \frac{5}{6} \right| \quad \Rightarrow \quad \frac{1}{6} (\hat{i} + 4\hat{j} + 7\hat{k}). \]