Given two lines defined by their Cartesian equations: $L_1: \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-

Question

Given two lines defined by their Cartesian equations: $L_1: \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}$ and $L_2: \frac{x+1}{2}=\frac{y-1}{\alpha}=\frac{z}{1}$. If the shortest distance separating these lines is $\frac{3}{\sqrt{14}}$, find the sum of all permissible values of $\alpha$.

Professor PC Thomas Classes · Answer

The given lines are $L_1: \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}$ and $L_2: \frac{x+1}{2}=\frac{y-1}{\alpha}=\frac{z}{1}$.
For line $L_1$, we can identify a point $A(2,-1,3)$ and its direction vector $\overrightarrow{b_1}=\hat{i}+2 \hat{j}-\hat{k}$.
For line $L_2$, we can identify a point $C(-1,1,0)$ and its direction vector $\overrightarrow{b_2}=2 \hat{i}+\alpha \hat{j}+\hat{k}$.
First, find the vector $\overrightarrow{A C}$ connecting a point on $L_1$ to a point on $L_2$ :
$$
\begin{aligned}
& \overrightarrow{A C}=C-A=(-1-2) \hat{i}+(1-(-1)) \hat{j}+(0-3) \hat{k}=-3 \hat{i}+2 \hat{j}-3 \hat{k} \
& \overrightarrow{b_1} 	imes \overrightarrow{b_2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \
1 & 2 & -1 \
2 & \alpha & 1
\end{array}ight| \
& =\hat{i}(2 \cdot 1-(-1) \cdot \alpha)-\hat{j}(1 \cdot 1-(-1) \cdot 2)+\hat{k}(1 \cdot \alpha-2 \cdot 2) \
& =(2+\alpha) \hat{i}-(1+2) \hat{j}+(\alpha-4) \hat{k} \
& =(2+\alpha) \hat{i}-3 \hat{j}+(\alpha-4) \hat{k} \
& (\overrightarrow{A C}) \cdot\left(\overrightarrow{b_1} 	imes \overrightarrow{b_2}ight)=(-3)(2+\alpha)+(2)(-3)+(-3)(\alpha-4) \
& =-6-3 \alpha-6-3 \alpha+12 \
& =-6 \alpha \
& \left|\overrightarrow{b_1} 	imes \overrightarrow{b_2}ight|=\sqrt{(2+\alpha)^2+(-3)^2+(\alpha-4)^2} \
& =\sqrt{\left(4+4 \alpha+\alpha^2ight)+9+\left(\alpha^2-8 \alpha+16ight)} \
& =\sqrt{2 \alpha^2-4 \alpha+29}
\end{aligned}
$$

The formula for the shortest distance (SD) between two skew lines is $S D=\frac{\left|(\overrightarrow{A C}) \cdot\left(\overrightarrow{b_1} 	imes \overrightarrow{b_2}ight)ight|}{\left|\overrightarrow{b_1} 	imes \overrightarrow{b_2}ight|}$.
We are given that the shortest distance is $\frac{3}{\sqrt{14}}$. So, $\frac{|-6 \alpha|}{\sqrt{2 \alpha^2-4 \alpha+29}}=\frac{3}{\sqrt{14}}$
$$
\begin{aligned}
& \frac{6|\alpha|}{\sqrt{2 \alpha^2-4 \alpha+29}}=\frac{3}{\sqrt{14}} \
& \frac{2|\alpha|}{\sqrt{2 \alpha^2-4 \alpha+29}}=\frac{1}{\sqrt{14}} \
& \frac{(2 \alpha)^2}{2 \alpha^2-4 \alpha+29}=\frac{1}{14} \
& \frac{4 \alpha^2}{2 \alpha^2-4 \alpha+29}=\frac{1}{14} \
& 14\left(4 \alpha^2ight)=1\left(2 \alpha^2-4 \alpha+29ight) \
& 56 \alpha^2=2 \alpha^2-4 \alpha+29 \
& 54 \alpha^2+4 \alpha-29=0
\end{aligned}
$$
We need to find the sum of all permissible values of $\alpha$. For a quadratic equation $a x^2+b x+c=0$, the sum of the roots is given by $-\frac{b}{a}$.
In this equation, $a=54, b=4$, and $c=-29$.
Sum of permissible values of $\alpha=-\frac{4}{54}=-\frac{2}{27}$.
The final answer is $-\frac{2}{27}$