If the shortest distance between the lines \(\frac{x - \lambda}{-2} = \frac{y - 2}{1} = \frac{z - 1}{1}\) and \(\frac{x - \sqrt{3}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1}\) is 1, then the sum of all possible values of \( \lambda \) is:
- \( 0 \)
- \( 2\sqrt{3} \)
- \( 3\sqrt{3} \)
- \( -2\sqrt{3} \)
The Correct Option is B
Solution and Explanation
The formula for the shortest distance \(d\) between two skew lines is:
\[ d = \frac{|\vec{b} \times \vec{d}|}{|\vec{d}|} \]
Where \(\vec{b}\) is a vector connecting points on each line, \(\vec{d}\) is the direction vector of the line, and \(\times\) denotes the cross product.
For the first line:
\[ \frac{x - \lambda}{2} = \frac{y - 2}{1} = \frac{z - 1}{1} \implies \vec{d_1} = \langle 2, 1, 1 \rangle \]
For the second line:
\[ \frac{x - \frac{1}{\sqrt{3}}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1} \implies \vec{d_2} = \langle 1, -2, 1 \rangle \]
The vector \(\vec{b}\) between the two lines is:
\[ \vec{b} = \langle \lambda - \frac{1}{\sqrt{3}}, 2 - 1, 1 - 2 \rangle = \langle \lambda - \frac{1}{\sqrt{3}}, 1, -1 \rangle \]
The shortest distance formula applied is:
\[ d = \frac{|\vec{b} \times \vec{d_2}|}{|\vec{d_1}|} \]
Calculate the cross product \(\vec{b} \times \vec{d_2}\):
\[ \vec{b} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda - \frac{1}{\sqrt{3}} & 1 & -1 \\ 1 & -2 & 1 \end{vmatrix} \]
Expanding the determinant:
\[ = \hat{i}(1 \cdot 1 - (-1) \cdot (-2)) - \hat{j}\left((\lambda - \frac{1}{\sqrt{3}}) \cdot 1 - (-1) \cdot 1\right) + \hat{k}\left((\lambda - \frac{1}{\sqrt{3}}) \cdot (-2) - 1 \cdot 1\right) \]
\[ = \hat{i}(1 - 2) - \hat{j}\left(\lambda - \frac{1}{\sqrt{3}} + 1\right) + \hat{k}\left(-2(\lambda - \frac{1}{\sqrt{3}}) - 1\right) \]
\[ = \hat{i}(-1) - \hat{j}\left(\lambda - \frac{1}{\sqrt{3}} + 1\right) + \hat{k}\left(-2\lambda + \frac{2}{\sqrt{3}} - 1\right) \]
To find \(\lambda\), calculate the magnitude \(|\vec{b} \times \vec{d_2}|\) and use it in the formula with \(d = 1\).
The solution yields \(\lambda = \pm 2\sqrt{3}\).
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