Question

Find the shortest distance between the lines $\frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}\mathrm{and}\frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}.$

Answer 1

$$\begin{gathered} \text{The given equations of the lines are} \\ \frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}...\left(1\right) \\ \frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}...\left(2\right) \\ \text{Clearly (2) passes through the point}\mathit{\text{P}}\text{(0, -1, 1).} \\ \text{Let the direction ratios of the plane be proportional to}a,b,c. \\ \text{Since the plane containing line (1) should pass through (2, 5, 0) and is parallel to the line (1),} \\ \text{equation of the plane passing through (1) is} \\ a\left(x-2\right)+b\left(y-5\right)+c\left(z-0\right)=0...\left(3\right), \\ \text{where -}a+2b+3c=0...\left(4\right) \\ \text{Since the plane is parallel to line (2),} \\ 2a-b+2c=0...\left(5\right) \\ \text{Solving (4) and (5) using cross-multiplication, we get} \\ \frac{a}{7}=\frac{b}{8}=\frac{c}{-3} \\ \text{Substituting}a,b\text{and}\mathit{\text{c}}\text{in (3), we get} \\ 7\left(x-2\right)+8\left(y-5\right)-3\left(z-0\right)=0 \\ \Rightarrow 7x+8y-3z-54=0...\left(6\right), \\ \text{which is the equation of the plane containing line (1) and parallel to line (2).} \\ \text{Shortest distance between (1) and (2)} \\ =\text{Distance between the point}\mathit{\text{P}}\text{(0, -1, 1) and plane (6)} \\ =\left|\frac{7\left(0\right)+8\left(-1\right)-3\left(1\right)-54}{\sqrt{49+64+9}}\right| \\ =\frac{65}{\sqrt{122}}\text{units} \end{gathered}$$