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 a systematic approach.

### Step 1: Determine the coordinates of point D
Given the coordinates of points A, B, and C:
- A(1, 1, 1)
- B(1, 2, 3)
- C(1, 1, 2)

The center of the tetrahedron is given as \((\frac{3}{2}, \frac{3}{4}, 2)\).

Using the formula for the centroid of a tetrahedron, we have:
\[ \frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{3}{2} \]
Substituting the known values:
\[ \frac{1 + 1 + 1 + x}{4} = \frac{3}{2} \]
Multiplying both sides by 4:
\[ 3 + x = 6 \implies x = 3 \]

Now for the y-coordinate:
\[ \frac{1 + 2 + 1 + y}{4} = \frac{3}{4} \]
Multiplying both sides by 4:
\[ 4 + y = 3 \implies y = -1 \]

Now for the z-coordinate:
\[ \frac{1 + 3 + 2 + z}{4} = 2 \]
Multiplying both sides by 4:
\[ 6 + z = 8 \implies z = 2 \]

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

### Step 2: Find the equations of lines AB and CD
**Line AB:**
Using the parametric form, the direction vector from A to B is:
\[ \vec{AB} = B - A = (1 - 1, 2 - 1, 3 - 1) = (0, 1, 2) \]
The equation of line AB can be written as:
\[ \frac{x - 1}{0} = \frac{y - 1}{1} = \frac{z - 1}{2} \]

**Line CD:**
The direction vector from C to D is:
\[ \vec{CD} = D - C = (3 - 1, -1 - 1, 2 - 2) = (2, -2, 0) \]
The equation of line CD can be written as:
\[ \frac{x - 1}{2} = \frac{y - 1}{-2} = \frac{z - 2}{0} \]

### Step 3: Use the formula for the shortest distance between two skew lines
The formula for the shortest distance \(d\) between two skew lines is given by:
\[ d = \frac{|(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})|}{|\vec{b_1} \times \vec{b_2}|} \]
Where:
- \(\vec{a_1} = A(1, 1, 1)\)
- \(\vec{b_1} = \vec{AB} = (0, 1, 2)\)
- \(\vec{a_2} = C(1, 1, 2)\)
- \(\vec{b_2} = \vec{CD} = (2, -2, 0)\)

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

Now, calculate \(|\vec{b_1} \times \vec{b_2}|\):
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \]

Now calculate \(\vec{a_2} - \vec{a_1}\):
\[ \vec{a_2} - \vec{a_1} = C - A = (1 - 1, 1 - 1, 2 - 1) = (0, 0, 1) \]

Now calculate \((\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})\):
\[ (4, 4, -2) \cdot (0, 0, 1) = 4 \cdot 0 + 4 \cdot 0 + (-2) \cdot 1 = -2 \]

Finally, substitute into the distance formula:
\[ d = \frac{| -2 |}{6} = \frac{2}{6} = \frac{1}{3} \]

### Final Answer
The shortest distance between the skew lines AB and CD is \(\frac{1}{3}\).