Class 12: The Plane – Exercise 29.14

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

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

$\displaystyle \textbf{Question 3: }~\text{Find the shortest distance between the lines }\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}\text{ and }3x-y-2z+4=0=2x+y+z+1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane containing the line}$
$\displaystyle 3x-y-2z+4=0 \text{ and } 2x+y+z+1 \text{ is}$
$\displaystyle (3x-y-2z+4)+\lambda(2x+y+z+1)=0$
$\displaystyle \text{Or }(3+2\lambda)x+(\lambda-1)y+(\lambda-2)z+(\lambda+4)=0\ldots(1)$
$\displaystyle \text{If it is parallel to the line}$
$\displaystyle \frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1},\text{ then}$
$\displaystyle 2(3+2\lambda)+4(\lambda-1)+(\lambda-2)=0$
$\displaystyle \Rightarrow 9\lambda=0$
$\displaystyle \Rightarrow \lambda=0$
$\displaystyle \text{Putting }\lambda=0\text{ in (1), we get}$
$\displaystyle 3x-y-2z+4=0\ldots(2)$
$\displaystyle \text{This is the equation of the plane containing the second line and parallel to the first line.}$
$\displaystyle \text{Now, the line }\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$
$\displaystyle \text{passes through }(1,3,-2).$
$\displaystyle \text{Shortest distance between the given lines}$
$\displaystyle =\text{Length of the perpendicular from }(1,3,-2)\text{ to the plane}$
$\displaystyle 3x-y-2z+4=0$
$\displaystyle =\frac{|3(1)-3-2(-2)+4|}{\sqrt{3^2+(-1)^2+(-2)^2}}$
$\displaystyle =\frac{|3-3+4+4|}{\sqrt{9+1+4}}$
$\displaystyle =\frac{8}{\sqrt{14}}\text{ units}$