Question

The shortest distance between the skew lines \( \vec{r} = (2\hat{i} - \hat{j}) + t(\hat{i} + 2\hat{k}) \) and \( \vec{r} = (-2\hat{i} + \hat{k}) + s(\hat{i} - \hat{j} - \hat{k}) \) is:

• A. \( \frac{3\sqrt{2}}{\sqrt{7}} \)
• B. \( \frac{3}{\sqrt{7}} \)
• C. \( \frac{3}{\sqrt{14}} \)
• D. \( \frac{4}{\sqrt{14}} \)

Hint:

To find the shortest distance between two skew lines, use the formula involving the cross product of direction vectors and the vector connecting points on the lines.

Answer 1

The Correct Option is A

We are given the parametric equations of two skew lines: \[ \vec{r}_1 = (2\hat{i} - \hat{j}) + t(\hat{i} + 2\hat{k}), \] and \[ \vec{r}_2 = (-2\hat{i} + \hat{k}) + s(\hat{i} - \hat{j} - \hat{k}). \] To find the shortest distance between these skew lines, we use the formula: \[ d = \frac{|(\vec{r}_2 - \vec{r}_1) \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}, \] where \( \vec{v}_1 \) and \( \vec{v}_2 \) are the direction vectors of the two lines, and \( \vec{r}_1 \) and \( \vec{r}_2 \) are points on the lines. Here, \[ \vec{v}_1 = \hat{i} + 2\hat{k}, \quad \vec{v}_2 = \hat{i} - \hat{j} - \hat{k}, \] and \[ \vec{r}_2 - \vec{r}_1 = (-2\hat{i} + \hat{k}) - (2\hat{i} - \hat{j}) = -4\hat{i} + \hat{j} + \hat{k}. \] Now, compute the cross product \( \vec{v}_1 \times \vec{v}_2 \), and then substitute in the formula for the shortest distance. After performing the necessary calculations, we get: \[ d = \frac{3\sqrt{2}}{\sqrt{7}}. \] Thus, the correct answer is \( \frac{3\sqrt{2}}{\sqrt{7}} \).