Question

If the shortest distance between the lines $\dfrac{x-1}{\alpha }=\dfrac{y+1}{-1}=\dfrac{z}{1},\left( \alpha \ne -1 \right)\ and\ x+y+z+1=0=2x-y+z+3\ is\ \dfrac{1}{\sqrt{3}}$ , then the value of $\alpha$ is:
A. $\dfrac{32}{19}$
B. $-\dfrac{16}{19}$
C. $-\dfrac{19}{16}$
D. $\dfrac{19}{32}$

Answer 1

First, assume x = 0, and find the value of y and z by putting x = 0 in the equation x + y + z + 1 = 0 and 2x – y + z + 3 = 0. Then, we will get the direction ratios of both the lines. After that use the given formula to find the shortest distance between the given lines and equate it to $\dfrac{1}{\sqrt{3}}$.
Shortest distance between two lines,
$\begin{aligned} & {{l}_{1}}:\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}} \\\ & {{l}_{2}}:\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}} \\\ & \dfrac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|}{\sqrt{{{\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}+{{\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)}^{2}}+{{\left( {{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}} \right)}^{2}}}} \\\ \end{aligned}$

Complete step-by-step answer:
Given, the shortest distance between the lines $\dfrac{x-1}{\alpha }=\dfrac{y+1}{-1}=\dfrac{z}{1}\ and\ x+y+z+1=0=2x-y+z+3\ is\ \dfrac{1}{\sqrt{3}}$.
Now, let us consider $x+y+z+1=0\ and\ 2x-y+z+3$.
Let us assume, the value of x = 0. Putting x = 0 in equation $x+y+z+1=0$, we will get,
$\begin{aligned} & 0+y+z+1=0 \\\ & \Rightarrow y+z+1=0.........\left( 1 \right) \\\ \end{aligned}$
Putting x = 0, in equation $2x-y+z+3$, we will get,
$\begin{aligned} & 2\left( 0 \right)-y+z+3=0 \\\ & \Rightarrow -y+z+3=0.........\left( 2 \right) \\\ \end{aligned}$
Now, we will solve equation (1) and (2).
Adding equation (1) and (2), we get,
$\begin{aligned} & \Rightarrow \left( y+z+1 \right)+\left( -y+z+3 \right)=0 \\\ & \Rightarrow y+z+1-y+z+3=0 \\\ & \Rightarrow 2z+4=0 \\\ & \Rightarrow 2z=\left( -4 \right) \\\ & \therefore z=\dfrac{-4}{2} \\\ & \therefore z=-2 \\\ \end{aligned}$
Putting $z=\left( -2 \right)$ in equation (1), we get,
$\begin{aligned} & \Rightarrow y+\left( -2 \right)+1=0 \\\ & \Rightarrow y-1=0 \\\ & \Rightarrow y=1 \\\ \end{aligned}$
For any line, $ax+by+cz+d=0$, a, b and c represents direction ratios of the line. Therefore, for given line $x+y+z+1=0$, line has direction ratios 1, 1, 1.
For the given line $2x-y+z+3=0$, line has direction ratios 2, -1, 1.
We know that line will be perpendicular when, a + b + c = 0 and 2a – b + c = 0.
So, using the cross multiplication method, we get,
$\begin{aligned} & \dfrac{a}{1+1}=\dfrac{b}{2-1}=\dfrac{c}{-1-2} \\\ & \Rightarrow \dfrac{a}{2}=\dfrac{b}{1}=\dfrac{c}{-3} \\\ \end{aligned}$
So, we get (2, 1, -3) as direction ratios.
Now, we can write this as,
$\begin{aligned} & \dfrac{x}{2}=\dfrac{y-1}{1}=\dfrac{z+2}{-3} \\\ & For, \\\ & \Rightarrow \dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}\ and \\\ & \Rightarrow \dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}} \\\ \end{aligned}$
By, applying the formula we get the shortest distance,
$\Delta =\dfrac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|}{\sqrt{{{\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}+{{\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)}^{2}}+{{\left( {{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}} \right)}^{2}}}}$
We have lines as $\dfrac{x-1}{\alpha }=\dfrac{y+1}{-1}=\dfrac{z}{1}$ and $\dfrac{x}{2}=\dfrac{y-1}{1}=\dfrac{z+2}{-3}$. Substituting the values in the above formula of shortest distance for the equations of given lines, we get,
$\Delta =\dfrac{\left| \begin{matrix} 0-1 & 1-(-1) & -2-0 \\\ \alpha & -1 & 1 \\\ 2 & 1 & -3 \\\ \end{matrix} \right|}{\sqrt{{{\left( \alpha +2 \right)}^{2}}+{{\left( 3-1 \right)}^{2}}+{{\left( 2+3\alpha \right)}^{2}}}}$

-1 & 2 & -2 \\\ \alpha & -1 & 1 \\\ 2 & 1 & -3 \\\ \end{matrix} \right|}{\sqrt{{{\left( \alpha +2 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( 2+3\alpha \right)}^{2}}}}$$ Now, we will expand the determinant and simplify the denominator using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . $$\begin{aligned} & \Delta =\dfrac{-1\left[ 3-1 \right]-2\left[ -3\alpha -2 \right]-2\left[ \alpha +2 \right]}{\sqrt{{{\alpha }^{2}}+4\alpha +4+4+4+12\alpha +9{{\alpha }^{2}}}} \\\ & \Rightarrow \Delta =\dfrac{-2+6\alpha +4-2\alpha -4}{\sqrt{10{{\alpha }^{2}}+16\alpha +12}} \\\ & \Rightarrow \Delta =\dfrac{4\alpha -2}{\sqrt{10{{\alpha }^{2}}+16\alpha +12}}.........\left( 1 \right) \\\ \end{aligned}$$ Now, equating the value of equation (1) with $\dfrac{1}{\sqrt{3}}$, shortest distance between the given lines, we get, $\begin{aligned} & \dfrac{1}{\sqrt{3}}=\dfrac{4\alpha -2}{\sqrt{10{{\alpha }^{2}}+16\alpha +12}} \\\ & \Rightarrow 3{{\left( 4\alpha -2 \right)}^{2}}=10{{\alpha }^{2}}+16\alpha +12 \\\ & \Rightarrow 3\left[ 16{{\alpha }^{2}}+4-16\alpha \right]=10{{\alpha }^{2}}+16\alpha +12 \\\ \end{aligned}$ On expanding the above equation, we get, $$\begin{aligned} & \Rightarrow 48{{\alpha }^{2}}+12-48\alpha =10{{\alpha }^{2}}+16\alpha +12 \\\ & \Rightarrow 48{{\alpha }^{2}}-10{{\alpha }^{2}}+12-12-48\alpha -16\alpha =0 \\\ & \Rightarrow 38{{\alpha }^{2}}-64\alpha =0 \\\ & \Rightarrow 38{{\alpha }^{2}}=64\alpha \\\ & \text{since, }\alpha \ne 0 \\\ & \therefore 38\alpha =64 \\\ & \Rightarrow \alpha =\dfrac{64}{38} \\\ \end{aligned}$$ Simplifying further ,we get $\alpha =\dfrac{32}{19}$ **Note:** In this equation, students may try to solve the equations of the given lines first. But in the question, we have given an equation of two lines in three variables. Hence, we need to assume the value of one either x, y or z to find the value of other variables. After that only, we can find the direction ratios of the given lines.