Vertices of a parallelogram taken in order are A, ( 2,-1,4) , B (1,0,-1) , C ( 1,2,3) and D (x,y,z) The distance between the parallel lines AB and CD is

Options

A. (a) `sqrt6`
B. (b) `3sqrt(6//5)`
C. (c) `2sqrt2`
D. (d) `3`

Explanation

To solve the problem of finding the distance between the parallel lines AB and CD of the parallelogram with vertices A(2, -1, 4), B(1, 0, -1), C(1, 2, 3), and D(x, y, z), we will follow these steps:

### Step 1: Find the coordinates of point D
Since the diagonals of a parallelogram bisect each other, we can find the coordinates of point D using the midpoint of the diagonals AC and BD.

1. **Find the midpoint O of diagonal AC:**
- A = (2, -1, 4)
- C = (1, 2, 3)
- Midpoint O = ((2 + 1)/2, (-1 + 2)/2, (4 + 3)/2) = (3/2, 1/2, 7/2)

2. **Set up equations for the midpoint of diagonal BD:**
- B = (1, 0, -1)
- D = (x, y, z)
- Midpoint O = ((1 + x)/2, (0 + y)/2, (-1 + z)/2)

3. **Equate the midpoints:**
- For x-coordinate: (1 + x)/2 = 3/2 → 1 + x = 3 → x = 2
- For y-coordinate: (0 + y)/2 = 1/2 → y = 1
- For z-coordinate: (-1 + z)/2 = 7/2 → -1 + z = 7 → z = 8

Thus, the coordinates of point D are D(2, 1, 8).

### Step 2: Find the vector AB and vector AD
1. **Vector AB:**
- A = (2, -1, 4)
- B = (1, 0, -1)
- Vector AB = B - A = (1 - 2, 0 + 1, -1 - 4) = (-1, 1, -5)

2. **Vector AD:**
- D = (2, 1, 8)
- Vector AD = D - A = (2 - 2, 1 + 1, 8 - 4) = (0, 2, 4)

### Step 3: Calculate the cross product of vectors AB and AD
Using the determinant method to find the cross product:

\[ \text{AB} \times \text{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -5 \\ 0 & 2 & 4 \end{vmatrix} \]

Calculating the determinant:
- i-component: \(1 \cdot 4 - (-5) \cdot 2 = 4 + 10 = 14\)
- j-component: \(-(-1 \cdot 4 - 0 \cdot -5) = -(-4) = 4\)
- k-component: \(-1 \cdot 2 - 1 \cdot 0 = -2\)

Thus, \(\text{AB} \times \text{AD} = (14, 4, -2)\).

### Step 4: Calculate the magnitude of vector AB
\[ |\text{AB}| = \sqrt{(-1)^2 + 1^2 + (-5)^2} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3} \]

### Step 5: Calculate the distance between the lines AB and CD
The distance \(d\) between the parallel lines is given by the formula:
\[ d = \frac{|\text{AB} \times \text{AD}|}{|\text{AB}|} \]

1. **Calculate the magnitude of the cross product:**
\[ |\text{AB} \times \text{AD}| = \sqrt{14^2 + 4^2 + (-2)^2} = \sqrt{196 + 16 + 4} = \sqrt{216} = 6\sqrt{6} \]

2. **Substituting into the distance formula:**
\[ d = \frac{6\sqrt{6}}{3\sqrt{3}} = 2\sqrt{2} \]

### Final Answer
The distance between the parallel lines AB and CD is \(2\sqrt{2}\).

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