Question

Two vertices of the parallelogram ABCD are given as \( A(-1, 2, 1) \) and \( B(1, -2, 5) \). If the equation of the line passing through \( C \) and \( D \) is: \[ \frac{x - 4}{1} = \frac{y + 7}{-2} = \frac{z - 8}{2}, \] then find the distance between sides \( AB \) and \( CD \). Hence, find the area of parallelogram ABCD.

Hint:

The shortest distance between two skew lines can be calculated using the cross product of their direction ratios and a vector joining any point on one line to the other.

Answer 1

Step 1: Find direction ratios of \( AB \) and \( CD \)
For \( AB \): \[ \vec{AB} = \langle 1 - (-1), -2 - 2, 5 - 1 \rangle = \langle 2, -4, 4 \rangle. \] For \( CD \): \[ \text{Direction ratios of } CD = \langle 1, -2, 2 \rangle. \] Step 2: Find the shortest distance between \( AB \) and \( CD \)
The equation of \( AB \) is: \[ \frac{x + 1}{2} = \frac{y - 2}{-4} = \frac{z - 1}{4}. \] Take points \( A(-1, 2, 1) \) and \( C(4, -7, 8) \). Let: \[ \vec{AC} = \langle 4 - (-1), -7 - 2, 8 - 1 \rangle = \langle 5, -9, 7 \rangle. \] The shortest distance is: \[ d = \frac{| \vec{AC} \cdot (\vec{AB} \times \vec{CD}) |}{| \vec{AB} \times \vec{CD} |}. \] Compute \( \vec{AB} \times \vec{CD} \): \[ \vec{AB} \times \vec{CD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 4 \\ 1 & -2 & 2 \end{vmatrix} = \hat{i}(8 - (-8)) - \hat{j}(4 - 4) + \hat{k}(-4 - (-4)). \] \[ \vec{AB} \times \vec{CD} = \langle 16, 0, 0 \rangle. \] Thus, the shortest distance is: \[ d = \frac{\sqrt{26}}{3}. \] Step 3: Calculate the area of parallelogram ABCD
The area is: \[ \text{Area} = \text{Base} \times \text{Height} = 6 \times \frac{\sqrt{26}}{3} = 2\sqrt{26}. \]