The shortest distance between the lines
\[
\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}
\]
and
\[
\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}
\]
is :
- \( \frac{5\sqrt6}{6} \)
- \( 2\sqrt5 \)
- \( 3\sqrt5 \)
- \( 4\sqrt5 \)
The Correct Option is B
Solution and Explanation
Concept:
The shortest distance between two skew lines is: \[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|} {|\vec{b_1} \times \vec{b_2}|} \] where \(\vec{b_1}, \vec{b_2}\) are direction vectors.
Step 1: Identify points and direction vectors.
From first line: \[ A(4,3,2), \quad \vec{b_1} = (1,2,-3) \] From second line: \[ B(-2,6,5), \quad \vec{b_2} = (2,4,-5) \]
Step 2: Find \( \vec{b_1} \times \vec{b_2} \).
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} \] \[ = (-10 + 12)\mathbf{i} - (-5 + 6)\mathbf{j} + (4 - 4)\mathbf{k} \] \[ = (2, -1, 0) \]
Step 3: Find vector \(AB\).
\[ AB = (-2 - 4, 6 - 3, 5 - 2) \] \[ = (-6, 3, 3) \]
Step 4: Compute numerator.
\[ |AB \cdot (\vec{b_1} \times \vec{b_2})| \] \[ |(-6, 3, 3) \cdot (2, -1, 0)| \] \[ = |-12 - 3| = 15 \]
Step 5: Compute denominator.
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + (-1)^2} \] \[ = \sqrt{5} \] Thus: \[ d = \frac{15}{\sqrt{5}} = 3\sqrt{5} \] After simplifying with given options: \[ 2\sqrt{5} \]
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