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: \[ \frac{x - 4}{1} = \frac{y + 7}{-2} = \frac{z - 8}{2}, \] find the distance between sides \( AB \) and \( CD \). Hence, find the area of the parallelogram \( ABCD \).

Hint:

To calculate the distance between skew lines, use the formula involving direction vectors and points on the lines. For area, multiply the distance by the magnitude of one side vector.

Answer 1

1. Vector representation of \( AB \): The position vectors of \( A \) and \( B \) are: \[ \vec{A} = \langle -1, 2, 1 \rangle, \quad \vec{B} = \langle 1, -2, 5 \rangle. \] The vector \( \overrightarrow{AB} \) is: \[ \overrightarrow{AB} = \vec{B} - \vec{A} = \langle 1 - (-1), -2 - 2, 5 - 1 \rangle = \langle 2, -4, 4 \rangle. \] 2. Direction vector of \( CD \): The direction vector of the line \( CD \) is: \[ \vec{d}_{CD} = \langle 1, -2, 2 \rangle. \] 3. Find a point on \( CD \): From the parametric equation of \( CD \): \[ x = 4 + t, \quad y = -7 - 2t, \quad z = 8 + 2t. \] At \( t = 0 \), a point on \( CD \) is: \[ C_0 = (4, -7, 8). \] 4. Shortest distance between \( AB \) and \( CD \): The formula for the shortest distance between skew lines is: \[ d = \frac{|(\vec{r}_2 - \vec{r}_1) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}. \] Here: - \( \vec{r}_1 = \langle -1, 2, 1 \rangle \) (point on \( AB \)), - \( \vec{r}_2 = \langle 4, -7, 8 \rangle \) (point on \( CD \)), - \( \vec{d}_1 = \overrightarrow{AB} = \langle 2, -4, 4 \rangle \), - \( \vec{d}_2 = \langle 1, -2, 2 \rangle \). Compute \( \vec{r}_2 - \vec{r}_1 \): \[ \vec{r}_2 - \vec{r}_1 = \langle 4 - (-1), -7 - 2, 8 - 1 \rangle = \langle 5, -9, 7 \rangle. \] Compute \( \vec{d}_1 \times \vec{d}_2 \): \[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -4 & 4
1 & -2 & 2 \end{vmatrix}. \] \[ \vec{d}_1 \times \vec{d}_2 = \hat{i} \begin{vmatrix} -4 & 4
-2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4
1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -4
1 & -2 \end{vmatrix}. \] \[ \vec{d}_1 \times \vec{d}_2 = \hat{i}(8 - (-8)) - \hat{j}(4 - 4) + \hat{k}(-4 - (-4)). \] \[ \vec{d}_1 \times \vec{d}_2 = \langle 16, 0, 0 \rangle. \] Compute \( (\vec{r}_2 - \vec{r}_1) \cdot (\vec{d}_1 \times \vec{d}_2) \): \[ (\vec{r}_2 - \vec{r}_1) \cdot (\vec{d}_1 \times \vec{d}_2) = \langle 5, -9, 7 \rangle \cdot \langle 16, 0, 0 \rangle = 5 \cdot 16 + 0 + 0 = 80. \] Compute \( |\vec{d}_1 \times \vec{d}_2| \): \[ |\vec{d}_1 \times \vec{d}_2| = \sqrt{16^2 + 0^2 + 0^2} = \sqrt{256} = 16. \] Shortest distance: \[ d = \frac{|80|}{16} = 5. \] 5. Area of the parallelogram: The area of the parallelogram is given by: \[ {Area} = |\overrightarrow{AB}| \cdot d, \] where: \[ |\overrightarrow{AB}| = \sqrt{2^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6. \] Substitute: \[ {Area} = 6 \cdot 5 = 30. \] Final Answer: The distance between \( AB \) and \( CD \) is \( \boxed{5} \), and the area of the parallelogram is \( \boxed{30} \).