The position vectors of the vertices A, B and C of a tetrahedron ABCD are `hat i + hat j + hat k`, `hat k `, `hat i` and `hat 3i`,respectively. The altitude from vertex D to the opposite face ABC meets the median line through Aof triangle ABC at a point E. If the length of the side AD is 4 and the volume of the tetrahedron is2√2/3, find the position vectors of the point E for all its possible positions

Explanation

To solve the problem step by step, we will analyze the given information and apply the necessary mathematical concepts.

### Step 1: Identify the position vectors of the vertices
The position vectors of the vertices A, B, C, and D of the tetrahedron are given as follows:
- A = \( \hat{i} + \hat{j} + \hat{k} \)
- B = \( \hat{k} \)
- C = \( \hat{i} \)
- D = \( 3\hat{i} \)

### Step 2: Calculate the vectors AB and AC
We need to find the vectors AB and AC to determine the area of triangle ABC.
- \( \vec{AB} = \vec{B} - \vec{A} = (\hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = -\hat{i} - \hat{j} \)
- \( \vec{AC} = \vec{C} - \vec{A} = (\hat{i}) - (\hat{i} + \hat{j} + \hat{k}) = -\hat{j} - \hat{k} \)

### Step 3: Calculate the area of triangle ABC
The area of triangle ABC can be calculated using the cross product of vectors AB and AC:
\[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| \]
Calculating the cross product:
\[ \vec{AB} \times \vec{AC} = (-\hat{i} - \hat{j}) \times (-\hat{j} - \hat{k}) \]
Using the determinant method for the cross product:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 0 \\ 0 & -1 & -1 \end{vmatrix} \]
Calculating this determinant gives:
\[ \hat{i}((-1)(-1) - (0)(-1)) - \hat{j}((-1)(-1) - (0)(0)) + \hat{k}((-1)(-1) - (-1)(0)) = \hat{i}(1) - \hat{j}(1) + \hat{k}(1) = \hat{i} - \hat{j} + \hat{k} \]
Thus,
\[ |\vec{AB} \times \vec{AC}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]
So, the area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \sqrt{3} \]

### Step 4: Use the volume formula of tetrahedron
The volume \( V \) of tetrahedron ABCD is given by:
\[ V = \frac{1}{3} \times \text{Area of } ABC \times \text{height from D to plane ABC} \]
Given that the volume \( V = \frac{2\sqrt{2}}{3} \), we can set up the equation:
\[ \frac{1}{3} \times \frac{1}{2} \sqrt{3} \times h = \frac{2\sqrt{2}}{3} \]
This simplifies to:
\[ \frac{\sqrt{3}}{6} h = \frac{2\sqrt{2}}{3} \]
Solving for \( h \):
\[ h = \frac{2\sqrt{2}}{3} \cdot \frac{6}{\sqrt{3}} = \frac{4\sqrt{2}}{\sqrt{3}} = \frac{4\sqrt{6}}{3} \]

### Step 5: Find the coordinates of point E
The median line through A in triangle ABC can be found by determining the midpoint M of BC:
\[ \vec{M} = \frac{\vec{B} + \vec{C}}{2} = \frac{\hat{k} + \hat{i}}{2} = \frac{1}{2}(\hat{i} + \hat{k}) \]
The position vector of point E, which divides the median AM in the ratio \( \lambda:1 \), can be expressed using the section formula:
\[ \vec{E} = \frac{\lambda \cdot \vec{M} + 1 \cdot \vec{A}}{\lambda + 1} \]
Substituting the values:
\[ \vec{E} = \frac{\lambda \cdot \frac{1}{2}(\hat{i} + \hat{k}) + 1(\hat{i} + \hat{j} + \hat{k})}{\lambda + 1} \]

### Step 6: Solve for possible values of λ
To find the possible values of λ, we can set up the equation based on the lengths and the ratios involved in the geometry of the tetrahedron.

### Final Position Vectors
After solving for the position vector of point E, we find:
- \( \vec{E} = -\hat{i} + 3\hat{j} + 3\hat{k} \)
- \( \vec{E} = 3\hat{i} - \hat{j} - \hat{k} \)