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

To solve this problem, we first need the centroid \( G \) of triangle \( ABC \), which acts as the intersection of the medians and divides them in a 2:1 ratio. Given the positions: 
A is \( \vec{A} = \hat{i} + 2\hat{j} + \hat{k} \),
B is \( \vec{B} = \hat{i} + 3\hat{j} - 2\hat{k} \),
C is \( \vec{C} = 2\hat{i} + \hat{j} - \hat{k} \).
We calculate the centroid: 
\[\vec{G} = \frac{1}{3}(\vec{A} + \vec{B} + \vec{C}) = \frac{1}{3}((\hat{i} + 2\hat{j} + \hat{k}) + (\hat{i} + 3\hat{j} - 2\hat{k}) + (2\hat{i} + \hat{j} - \hat{k}))\]
\[= \frac{1}{3}(4\hat{i} + 6\hat{j} - 2\hat{k}) = \frac{4}{3}\hat{i} + 2\hat{j} - \frac{2}{3}\hat{k}\]
Next, let's introduce the point \( D \), where the length \( \text{AD} = \frac{\sqrt{10}}{3} \). The volume \( V \) of the tetrahedron is \( \frac{\sqrt{805}}{6\sqrt{2}} \). The formula for the volume of a tetrahedron is: \[V = \frac{1}{3}\times \text{Area of base}\times \text{Height}\]
For the tetrahedron: 
Area of base \( ABC \) = \(|\vec{AB}\times \vec{AC}|\),
Let \(\vec{AB} = \vec{B} - \vec{A} = (0\hat{i} + \hat{j} - 3\hat{k})\)
\(\vec{AC} = \vec{C} - \vec{A} = (\hat{i} - \hat{j} - 2\hat{k})\)
Use the cross-product: 
\(\vec{AB} \times \vec{AC} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\0 & 1 & -3 \\1 & -1 & -2\end{vmatrix}\) = \((\hat{i}(3+2)-\hat{j}(0-3)+\hat{k}(0-1))\)
\[= 5\hat{i} + 3\hat{j} - \hat{k}\],
Magnitude = \(\sqrt{5^2 + 3^2 + 1^2} = \sqrt{35}\)
Area = \(\frac{1}{2}\times \sqrt{35}\).
Volume equation: \[\frac{1}{3} \times \frac{\sqrt{35}}{2} \times \text{Height} = \frac{\sqrt{805}}{6\sqrt{2}}\],
Solve for Height: \[\frac{\text{Height}}{3} = \frac{\sqrt{805} \times 2}{\sqrt{35} \times 2\sqrt{2}},\] \[\text{Height}= \sqrt{10}/3\],
which confirms the altitude aligns with \( \text{AD} \).
Since \( E \) is on the median, it divides it by a 2:1 ratio:
\[\text{Position vector of } E = \frac{1}{3}(2\vec{G} + \vec{A}) = \frac{1}{3}(2(\frac{4}{3}\hat{i} + 2\hat{j} - \frac{2}{3}\hat{k}) + (\hat{i} + 2\hat{j} + \hat{k}))\]
\(= \frac{1}{3}(\frac{8}{3}\hat{i} + 4\hat{j} - \frac{4}{3}\hat{k} + \hat{i} + 2\hat{j} + \hat{k})\)
\(= \frac{1}{3}(\frac{11}{3}\hat{i} + 6\hat{j} + \frac{2}{3}\hat{k})\)
\(= \frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k})\)
Therefore, the correct choice is: \( \frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k}) \).