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?
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?
Show Hint
- $\alpha+2 y=24$
- $2 \alpha+\gamma=7$
- $\alpha-2 \gamma=19$
- $2 \alpha-\gamma=9$
The Correct Option is A
Solution and Explanation
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 \( \lambda \).
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Concepts Used:
Shortest Distance Between Two Parallel Lines
Formula to find distance between two parallel line:
Consider two parallel lines are shown in the following form :
\(y = mx + c_1\) …(i)
\(y = mx + c_2\) ….(ii)
Here, m = slope of line
Then, the formula for shortest distance can be written as given below:
\(d= \frac{|c_2-c_1|}{\sqrt{1+m^2}}\)
If the equations of two parallel lines are demonstrated in the following way :
\(ax + by + d_1 = 0\)
\(ax + by + d_2 = 0\)
then there is a little change in the formula.
\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)