Question

Let the shortest distance between the lines $L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$ and $L_1: x+1=y-1=4-z$ be $2 \sqrt{6}$ If $(\alpha, \beta, \gamma)$ lies on $L$, then which of the following is NOT possible?

• A. $\alpha+2 y=24$
• B. $2 \alpha+\gamma=7$
• C. $\alpha-2 \gamma=19$
• D. $2 \alpha-\gamma=9$

Hint:

When solving for distances and coordinates, always check the signs and the conditions set by the system, and ensure the consistency of the equations.

Answer 1

The Correct Option is A

Step 1: The shortest distance between the lines is given by: \[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} = -\hat{i} - \hat{j} - 2\hat{k} \] This is the cross product of the direction vectors \( \mathbf{b_1} \) and \( \mathbf{b_2} \). Now, compute \( \mathbf{a_2} - \mathbf{a_1} \): \[ \mathbf{a_2} - \mathbf{a_1} = 6\hat{i} + (\lambda - 1)\hat{j} + (\lambda - 4)\hat{k} \] Step 2: The distance formula gives: \[ 2\sqrt{6} = \left| \frac{6 - \lambda + 1 + 2\lambda + 8}{\sqrt{1 + 1 + 4}} \right| \] Simplifying the equation: \[ | \lambda + 3 | = 12 \quad \Rightarrow \quad \lambda = 9, -15 \] The solutions for \( \lambda \) are 9 and -15. 
Step 3: From this, we find: \[ \alpha = -2k + 5, \quad \gamma = k - \lambda \quad \text{where} \quad k \in \mathbb{R} \] Thus, the equation becomes: \[ \alpha + 2\gamma = -13 \quad \text{or} \quad 35 \] This gives the final result for \( \alpha + 2\gamma \), which can be either -13 or 35, depending on the value of \( \lambda \).