Question

Find the shortest distance between the skew lines $\vec{r} = (-\hat{i} - 2\hat{j} - 3\hat{k}) + t(3\hat{i} - 2\hat{j} - 2\hat{k})$ and $\vec{r} = (7\hat{i} + 4\hat{k}) + s(\hat{i} - 2\hat{j} + 2\hat{k})$.

• A. \( 15 \)
• B. \( 0 \)
• C. \( 9 \)
• D. \( 16 \)

Hint:

o find the shortest distance between skew lines, use the formula: \( d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \) where $\vec{a_1}$ and $\vec{a_2}$ are points on the lines, and $\vec{b_1}$ and $\vec{b_2}$ are the direction vectors.

Answer 1

The Correct Option is C

Step 1: Identify the vectors.
Let the lines be \(\vec{r} = \vec{a_1} + t\vec{b_1}\) and \(\vec{r} = \vec{a_2} + s\vec{b_2}\), where:
 

• \(\vec{a_1} = -\hat{i} - 2\hat{j} - 3\hat{k}\)
• \(\vec{b_1} = 3\hat{i} - 2\hat{j} - 2\hat{k}\)
• \(\vec{a_2} = 7\hat{i} + 4\hat{k}\)
• \(\vec{b_2} = \hat{i} - 2\hat{j} + 2\hat{k}\)

 

Step 2: Calculate \(\vec{a_2} - \vec{a_1}\)
\(\vec{a_2} - \vec{a_1} = (7\hat{i} + 4\hat{k}) - (-\hat{i} - 2\hat{j} - 3\hat{k}) = 8\hat{i} + 2\hat{j} + 7\hat{k}\)

Step 3: Calculate \(\vec{b_1} \times \vec{b_2}\).
\(\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & -2 \\ 1 & -2 & 2 \end{vmatrix} = \hat{i}(-4 - 4) - \hat{j}(6 + 2) + \hat{k}(-6 + 2) = -8\hat{i} - 8\hat{j} - 4\hat{k}\)

Step 4: Find the magnitude of \(\vec{b_1} \times \vec{b_2}\).
\(|\vec{b_1} \times \vec{b_2}| = \sqrt{(-8)^2 + (-8)^2 + (-4)^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12\)

Step 5: Calculate the shortest distance.
The shortest distance d is given by:
\( d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| \)
\( d = \left| \frac{(8\hat{i} + 2\hat{j} + 7\hat{k}) \cdot (-8\hat{i} - 8\hat{j} - 4\hat{k})}{12} \right| \)
\( d = \left| \frac{-64 - 16 - 28}{12} \right| = \left| \frac{-108}{12} \right| = |-9| = 9 \)
Therefore, the shortest distance between the skew lines is 9.