Find the shortest distance between the lines \(\frac{x}{-1}=\frac{y-2}{0}=\frac{z}{1}\) and \(\frac{x+2}{1}=\frac{y}{1}=\frac{z}{0}\)

Concept:

The shortest distance between the lines \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\)  and \(\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}\) is given by:\(d= \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} & c_{1}\\ a_{2}& b_{2} & c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}\)

Calculation:

Here we have to find the shortest distance between the lines ​​\(\frac{x}{-1}=\frac{y-2}{0}=\frac{z}{1}\) and \(\frac{x+2}{1}=\frac{y}{1}=\frac{z}{0}\)

Let line L₁ be represented by the equation \(\frac{x}{-1}=\frac{y-2}{0}=\frac{z}{1}\) and line L₂ be represented by the equation \(\frac{x+2}{1}=\frac{y}{1}=\frac{z}{0}\)

⇒ x1 = 0, y1 = 2, z1 = 0  and a1 = -1, b1 = 0, c1 = 1.

⇒ x2 = -2, y2 = 0, z2 = 0  and a2 = 1, b2 = 1, c2 = 0.

∵ The shortest distance between the lines is given by:  \(d= \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} & c_{1}\\ a_{2}& b_{2} & c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}\)

⇒ \(d= \frac{\begin{vmatrix} -2-0 &0-2&0-0 \\ -1& 0 & 1\\ 1& 1 & 0 \end{vmatrix}}{\sqrt{(0-1)^{2}+(0-1)^{2}+(-1-0)^{2}}}\)    

⇒ \(d= \frac{\begin{vmatrix} -2 &-2&0 \\ -1& 0 & 1\\ 1& 1 & 0 \end{vmatrix}}{\sqrt{1+1+1}}\)

⇒ d = 0

Hence, option 4 is correct.