Consider a tetrahedron `D-ABC` with position vectors if its angular points as
A(1, 1, 1), B(1, 2, 3), C(1, 1, 2)
and centre of tetrahedron `((3)/(2), (3)/(4),2)`.
Q. Shortest distance between the skew lines AB and CD :

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(1)/(4)`
D. `(1)/(5)`

Answer

To find the shortest distance between the skew lines AB and CD in the tetrahedron D-ABC, we will follow these steps:

### Step 1: Determine the coordinates of point D
The center of the tetrahedron is given by the formula:
\[ \text{Center} = \left( \frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{y_1 + y_2 + y_3 + y_4}{4}, \frac{z_1 + z_2 + z_3 + z_4}{4} \right) \]
Given points:
- A(1, 1, 1)
- B(1, 2, 3)
- C(1, 1, 2)
- Center \(\left( \frac{3}{2}, \frac{3}{4}, 2 \right)\)

Let the coordinates of D be \((x_1, y_1, z_1)\).

Using the center coordinates, we can set up the equations:

1. For x-coordinates:
\[ \frac{1 + 1 + 1 + x_1}{4} = \frac{3}{2} \implies 3 + x_1 = 6 \implies x_1 = 3 \]

2. For y-coordinates:
\[ \frac{1 + 2 + 1 + y_1}{4} = \frac{3}{4} \implies 4 + y_1 = 3 \implies y_1 = -1 \]

3. For z-coordinates:
\[ \frac{1 + 3 + 2 + z_1}{4} = 2 \implies 6 + z_1 = 8 \implies z_1 = 2 \]

Thus, the coordinates of D are \(D(3, -1, 2)\).

### Step 2: Find the equations of lines AB and CD
**Line AB:**
Using points A(1, 1, 1) and B(1, 2, 3):
- Direction ratios of AB: \(B - A = (1-1, 2-1, 3-1) = (0, 1, 2)\)
- Parametric equations of line AB:
\[ \mathbf{r}_{AB} = (1, 1, 1) + \lambda(0, 1, 2) = (1, 1 + \lambda, 1 + 2\lambda) \]

**Line CD:**
Using points C(1, 1, 2) and D(3, -1, 2):
- Direction ratios of CD: \(D - C = (3-1, -1-1, 2-2) = (2, -2, 0)\)
- Parametric equations of line CD:
\[ \mathbf{r}_{CD} = (1, 1, 2) + \mu(2, -2, 0) = (1 + 2\mu, 1 - 2\mu, 2) \]

### Step 3: Find the shortest distance between the two lines
The formula for the shortest distance \(D\) between two skew lines is given by:
\[ D = \frac{|(\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
Where:
- \(\mathbf{b_1} = (0, 1, 2)\) (direction ratios of line AB)
- \(\mathbf{b_2} = (2, -2, 0)\) (direction ratios of line CD)
- \(\mathbf{a_1} = (1, 1, 1)\) (point A)
- \(\mathbf{a_2} = (1, 1, 2)\) (point C)

**Calculate \(\mathbf{b_1} \times \mathbf{b_2}\):**
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 2 \\ 2 & -2 & 0 \end{vmatrix} = \mathbf{i}(1 \cdot 0 - 2 \cdot (-2)) - \mathbf{j}(0 \cdot 0 - 2 \cdot 2) + \mathbf{k}(0 \cdot (-2) - 1 \cdot 2) \]
\[ = 0 + 4\mathbf{j} - 2\mathbf{k} = (0, 4, -2) \]

**Magnitude of \(\mathbf{b_1} \times \mathbf{b_2}\):**
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{0^2 + 4^2 + (-2)^2} = \sqrt{0 + 16 + 4} = \sqrt{20} = 2\sqrt{5} \]

**Calculate \(\mathbf{a_2} - \mathbf{a_1}\):**
\[ \mathbf{a_2} - \mathbf{a_1} = (1, 1, 2) - (1, 1, 1) = (0, 0, 1) \]

**Dot product \((\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1})\):**
\[ (0, 4, -2) \cdot (0, 0, 1) = 0 \cdot 0 + 4 \cdot 0 + (-2) \cdot 1 = -2 \]

**Calculate the shortest distance \(D\):**
\[ D = \frac{| -2 |}{2\sqrt{5}} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \]

### Final Answer
The shortest distance between the skew lines AB and CD is \(\frac{\sqrt{5}}{5}\).

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