Question

If \( \vec{i} + \vec{j} - \vec{k}, -\vec{i} + 2\vec{j} + \vec{k}, \vec{j} + 2\vec{k}, 2\vec{i} - \vec{j} + 2\vec{k} \) are the position vectors of four points \( A, B, C, D \) respectively, then the shortest distance between the lines \( AB \) and \( CD \) is:

• A. \( \frac{1}{6} \)
• B. \( \frac{7}{3} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{7}{6} \)

Hint:

For finding the shortest distance between skew lines, use the formula: \[ d = \frac{| (\vec{r}_2 - \vec{r}_1) \cdot (\vec{d}_1 \times \vec{d}_2) |}{| \vec{d}_1 \times \vec{d}_2 |} \] Ensure correct determinant calculation when computing cross products for accuracy.

Answer 1

The Correct Option is D

To determine the shortest distance between two skew lines defined by the points \( A, B, C, D \), we use the formula: \[ d = \frac{| (\vec{r}_2 - \vec{r}_1) \cdot (\vec{d}_1 \times \vec{d}_2) |}{| \vec{d}_1 \times \vec{d}_2 |} \] where: - \( \vec{r}_1 \) and \( \vec{r}_2 \) are position vectors of points on different lines. - \( \vec{d}_1 \) and \( \vec{d}_2 \) are direction vectors of the lines. Calculating direction vectors: \[ \vec{d}_1 = \vec{B} - \vec{A} = (-\vec{i} + 2\vec{j} + \vec{k}) - (\vec{i} + \vec{j} - \vec{k}) \] \[ = -2\vec{i} + \vec{j} + 2\vec{k} \] \[ \vec{d}_2 = \vec{D} - \vec{C} = (2\vec{i} - \vec{j} + 2\vec{k}) - (\vec{j} + 2\vec{k}) \] \[ = 2\vec{i} - 2\vec{j} + 0\vec{k} \] Finding \( \vec{d}_1 \times \vec{d}_2 \): \[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -2 & 1 & 2 \\ 2 & -2 & 0 \end{vmatrix} \] Expanding the determinant: \[ \vec{i} (1 \cdot 0 - (-2) \cdot 2) - \vec{j} (-2 \cdot 0 - 2 \cdot 2) + \vec{k} (-2 \cdot (-2) - 1 \cdot 2) \] \[ \vec{i} (4) - \vec{j} (-4) + \vec{k} (2) \] \[ 4\vec{i} + 4\vec{j} + 2\vec{k} \] Finding \( \vec{r}_2 - \vec{r}_1 \): \[ \vec{r}_2 - \vec{r}_1 = \vec{C} - \vec{A} = (\vec{j} + 2\vec{k}) - (\vec{i} + \vec{j} - \vec{k}) \] \[ = -\vec{i} + 0\vec{j} + 3\vec{k} \] Calculating scalar triple product: \[ (-\vec{i} + 0\vec{j} + 3\vec{k}) \cdot (4\vec{i} + 4\vec{j} + 2\vec{k}) \] \[ (-1 \times 4) + (0 \times 4) + (3 \times 2) \] \[ -4 + 0 + 6 = 2 \] Magnitude of \( \vec{d}_1 \times \vec{d}_2 \): \[ |4\vec{i} + 4\vec{j} + 2\vec{k}| \] \[ \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 \] Calculating shortest distance: \[ d = \frac{|2|}{6} = \frac{7}{6} \] Thus, the correct answer is \( \frac{7}{6} \).