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 \( \frac{3\sqrt{2}}{\sqrt{7}} \).

Answer 2

Given two skew lines:
\[ \vec{r}_1 = (2\hat{i} - \hat{j}) + t(\hat{i} + 2\hat{k}) \] \[ \vec{r}_2 = (-2\hat{i} + \hat{k}) + s(\hat{i} - \hat{j} - \hat{k}) \] Find the shortest distance between them.

Step 1: Identify direction vectors and points on each line:
\[ \vec{d}_1 = \hat{i} + 0\hat{j} + 2\hat{k} = (1, 0, 2) \] \[ \vec{d}_2 = \hat{i} - \hat{j} - \hat{k} = (1, -1, -1) \] Points on lines:
\[ \vec{A} = (2, -1, 0), \quad \vec{B} = (-2, 0, 1) \]

Step 2: The shortest distance \( D \) between skew lines is given by:
\[ D = \frac{|(\vec{B} - \vec{A}) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|} \]

Step 3: Calculate \( \vec{B} - \vec{A} \):
\[ \vec{B} - \vec{A} = (-2 - 2, 0 - (-1), 1 - 0) = (-4, 1, 1) \]

Step 4: Calculate cross product \( \vec{d}_1 \times \vec{d}_2 \):
\[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ 1 & -1 & -1 \\ \end{vmatrix} = \hat{i}(0 \times -1 - 2 \times -1) - \hat{j}(1 \times -1 - 2 \times 1) + \hat{k}(1 \times -1 - 0 \times 1) \] \[ = \hat{i}(0 + 2) - \hat{j}(-1 - 2) + \hat{k}(-1 - 0) = 2 \hat{i} + 3 \hat{j} - \hat{k} = (2, 3, -1) \]

Step 5: Calculate numerator \( |(\vec{B} - \vec{A}) \cdot (\vec{d}_1 \times \vec{d}_2)| \):
\[ (-4, 1, 1) \cdot (2, 3, -1) = (-4)(2) + (1)(3) + (1)(-1) = -8 + 3 - 1 = -6 \] Absolute value:
\[ | -6 | = 6 \]

Step 6: Calculate denominator \( |\vec{d}_1 \times \vec{d}_2| \):
\[ \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

Step 7: Shortest distance:
\[ D = \frac{6}{\sqrt{14}} = \frac{3 \times 2}{\sqrt{14}} = \frac{3 \sqrt{2}}{\sqrt{7}} \] (since \( \sqrt{14} = \sqrt{7 \times 2} = \sqrt{7} \sqrt{2} \), divide numerator and denominator by \( \sqrt{2} \))

Therefore,
\[ \boxed{\frac{3 \sqrt{2}}{\sqrt{7}}} \]