Question

Find the shortest distance between the pair of lines:
Line \( L_1 \) : \[ \frac{x - 1}{2} = \frac{y + 1}{3} = z, \] Line \( L_2 \) : \[ \frac{x + 1}{5} = \frac{y - 2}{1} = z = 2. \]

Answer 1

Question:
Find the shortest distance between the pair of lines:
Line \( L_1 \) : \[ \frac{x - 1}{2} = \frac{y + 1}{3} = z, \] Line \( L_2 \) : \[ \frac{x + 1}{5} = \frac{y - 2}{1} = z = 2. \]

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Solution:
Step 1: Write both lines in vector form:
Line \( L_1 \): \( \vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1 \), where
\( \vec{a}_1 = (1, -1, 0) \), \( \vec{b}_1 = (2, 3, 1) \).
Line \( L_2 \): \( \vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2 \).
From the line equations, since \( z = 2 \), plug into \( x \) and \( y \):
\( x = -1 + 5z = 9 \), \( y = 2 + z = 4 \), so \( \vec{a}_2 = (9, 4, 2) \).
\( \vec{b}_2 = (5, 1, 1) \).
Step 2: Use the formula:
\[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} \]
Step 3: Compute \( \vec{a}_2 - \vec{a}_1 = (8, 5, 2) \).
Step 4: Compute the cross product \( \vec{b}_1 \times \vec{b}_2 \):
\[ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 1 \\ \end{vmatrix} = \hat{i}(3 \cdot 1 - 1 \cdot 1) - \hat{j}(2 \cdot 1 - 5 \cdot 1) + \hat{k}(2 \cdot 1 - 15) = (2, 3, -13) \]
Step 5: Compute the dot product:
\[ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (8, 5, 2) \cdot (2, 3, -13) = 8 \cdot 2 + 5 \cdot 3 + 2 \cdot (-13) = 16 + 15 - 26 = 5 \]
Step 6: Compute the magnitude:
\[ |\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + 3^2 + (-13)^2} = \sqrt{4 + 9 + 169} = \sqrt{182} \]
Final Answer:
\[ \text{Shortest distance} = \frac{5}{\sqrt{182}} \]