Question

Find the shortest distance between the lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\mathrm{and}\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}.$

Answer 1

$$\begin{gathered} \text{The given equations of the lines are} \\ \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}...\left(1\right) \\ \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}...\left(2\right) \\ \text{Clearly (2) passes through the point}\mathit{\text{P}}\text{(3, 5, 7).} \\ \text{Let the direction ratios of the plane be proportional to}a,b,c. \\ \text{Since the plane contains line (1), it should pass through (-1, -1, -1) and is parallel to the line (1).} \\ \text{Equation of the plane through (1) is} \\ a\left(x+1\right)+b\left(y+1\right)+c\left(z+1\right)=0...\left(3\right), \\ \text{where 7}a-6b+c=0...\left(4\right) \\ \text{Since the plane is parallel to the line (2),} \\ a-2b+c=0...\left(5\right) \\ \text{Solving (4) and (5) using cross-multiplication, we get} \\ \frac{a}{-4}=\frac{b}{-6}=\frac{c}{-8} \\ \Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{4} \\ \text{Substituting}a,b\text{and}\mathit{\text{c}}\text{in (3), we get} \\ 2\left(x+1\right)+3\left(y+1\right)+4\left(z+1\right)=0 \\ \Rightarrow 2x+3y+4z+9=0...\left(6\right) \\ \text{which is the equation of the plane containing line (1) and parallel to line (2).} \\ \text{Shortest distance between (1) and (2)} \\ =\text{Distance between the point}\mathit{\text{P}}(3,\; 5,\; 7)\text{and plane (6)} \\ =\left|\frac{2\left(3\right)+3\left(5\right)+4\left(7\right)+9}{\sqrt{4+9+16}}\right| \\ =\frac{58}{\sqrt{29}} \\ =2\sqrt{29}\text{units} \end{gathered}$$