Question

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:

• A. \( 0 \)
• B. \( 2\sqrt{3} \)
• C. \( 3\sqrt{3} \)
• D. \( -2\sqrt{3} \)

Answer 1

The Correct Option is B

The shortest distance \(d\) between two skew lines is given by the formula:

\[ d = \frac{|\vec{b} \times \vec{d}|}{|\vec{d}|} \]

Where: \(\vec{b}\) is the vector joining points on each line, \(\vec{d}\) is the direction vector of the line, and \(\times\) represents 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:

\[ \frac{x - \frac{1}{\sqrt{3}}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1} \implies \vec{d_2} = \langle 1, -2, 1 \rangle \]

Now, the vector \(\vec{b}\) between the two lines can be written as:

\[ \vec{b} = \langle \lambda - \frac{1}{\sqrt{3}}, 2 - 1, 1 - 2 \rangle = \langle \lambda - \frac{1}{\sqrt{3}}, 1, -1 \rangle \]

The shortest distance formula becomes:

\[ d = \frac{|\vec{b} \times \vec{d_2}|}{|\vec{d_1}|} \]

Compute 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 this determinant:

\[ = \hat{i}(1 \cdot 1 - (-1) \cdot (-2)) - \hat{j}\left((\lambda - \frac{1}{\sqrt{3}}) \cdot 1 - (-1) \cdot 1\right) + \hat{k}\left((\lambda - \frac{1}{\sqrt{3}}) \cdot (-2) - 1 \cdot 1\right) \]

\[ = \hat{i}(1 - 2) - \hat{j}\left(\lambda - \frac{1}{\sqrt{3}} + 1\right) + \hat{k}\left(-2(\lambda - \frac{1}{\sqrt{3}}) - 1\right) \]

\[ = \hat{i}(-1) - \hat{j}\left(\lambda - \frac{1}{\sqrt{3}} + 1\right) + \hat{k}\left(-2\lambda + \frac{2}{\sqrt{3}} - 1\right) \]

Now, calculate the magnitude \(|\vec{b} \times \vec{d_2}|\) and use it in the formula for \(d = 1\) to solve for \(\lambda\).

After solving, we get \(\lambda = \pm 2\sqrt{3}\).