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
Approach Solution - 1
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 \(\lambda = \pm 2\sqrt{3}\).
Approach Solution -2
Given two lines:
L₁: \(\frac{x - \lambda}{-2} = \frac{y - 2}{1} = \frac{z - 1}{1}\)
L₂: \(\frac{x - \sqrt{3}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1}\)
The shortest distance between them is 1. We need to find the sum of all possible values of \(\lambda\).
Concept Used:
The shortest distance between two skew lines \(\vec{r} = \vec{a_1} + t\vec{b_1}\) and \(\vec{r} = \vec{a_2} + s\vec{b_2}\) is given by:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]
Step-by-Step Solution:
Step 1: Identify vectors from the given lines.
For L₁: Point \(A_1 = (\lambda, 2, 1)\), Direction vector \(\vec{b_1} = (-2, 1, 1)\)
For L₂: Point \(A_2 = (\sqrt{3}, 1, 2)\), Direction vector \(\vec{b_2} = (1, -2, 1)\)
Step 2: Compute \(\vec{A_1A_2} = \vec{a_2} - \vec{a_1}\).
\[ \vec{A_1A_2} = (\sqrt{3} - \lambda, 1 - 2, 2 - 1) = (\sqrt{3} - \lambda, -1, 1) \]
Step 3: Compute \(\vec{b_1} \times \vec{b_2}\).
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 1 \cdot (-2)) - \hat{j}((-2) \cdot 1 - 1 \cdot 1) + \hat{k}((-2) \cdot (-2) - 1 \cdot 1) \] \[ = \hat{i}(1 + 2) - \hat{j}(-2 - 1) + \hat{k}(4 - 1) = (3, 3, 3) \]
So, \(\vec{b_1} \times \vec{b_2} = (3, 3, 3)\)
Step 4: Compute the magnitude \(|\vec{b_1} \times \vec{b_2}|\).
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3} \]
Step 5: Compute the scalar triple product \((\vec{A_1A_2}) \cdot (\vec{b_1} \times \vec{b_2})\).
\[ (\vec{A_1A_2}) \cdot (\vec{b_1} \times \vec{b_2}) = (\sqrt{3} - \lambda, -1, 1) \cdot (3, 3, 3) \] \[ = 3(\sqrt{3} - \lambda) + 3(-1) + 3(1) = 3\sqrt{3} - 3\lambda - 3 + 3 = 3\sqrt{3} - 3\lambda \]
Step 6: Apply the distance formula and set it equal to 1.
\[ d = \frac{|3\sqrt{3} - 3\lambda|}{3\sqrt{3}} = 1 \] \[ \Rightarrow \frac{3|\sqrt{3} - \lambda|}{3\sqrt{3}} = 1 \] \[ \Rightarrow \frac{|\sqrt{3} - \lambda|}{\sqrt{3}} = 1 \] \[ \Rightarrow |\sqrt{3} - \lambda| = \sqrt{3} \]
Step 7: Solve the equation \(|\sqrt{3} - \lambda| = \sqrt{3}\).
\[ \sqrt{3} - \lambda = \sqrt{3} \quad \text{or} \quad \sqrt{3} - \lambda = -\sqrt{3} \] \[ \Rightarrow -\lambda = 0 \quad \text{or} \quad -\lambda = -2\sqrt{3} \] \[ \Rightarrow \lambda = 0 \quad \text{or} \quad \lambda = 2\sqrt{3} \]
Step 8: Find the sum of all possible values of \(\lambda\).
\[ \text{Sum} = 0 + 2\sqrt{3} = 2\sqrt{3} \]
Hence, the sum of all possible values of \(\lambda\) is \(2\sqrt{3}\).
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