Question

Equation of the line of shortest distace between the lines $\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$ and $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{5}$ is

• A. $3(x-21)=3y-92=3z-32$
• B. $\frac{x-\left(\frac{62}{3}\right)}{\frac{1}{3}}=\frac{y-31}{\frac{1}{3}}=\frac{z+\left(\frac{31}{3}\right)}{\frac{1}{3}}$
• C. $\frac{x-21}{\frac{1}{3}}=\frac{y-\left(\frac{92}{3}\right)}{\frac{1}{3}}=\frac{z+\left(\frac{32}{3}\right)}{\frac{1}{3}}$
• D. $\frac{x-2}{\frac{1}{3}}=\frac{y+3}{\frac{1}{3}}=\frac{z-1}{\frac{1}{3}}$

Answer 1

The correct option is A $3(x-21)=3y-92=3z-32$

Let $P(2r_1,-3r_1,r_1)$ and $Q(3r_2+2,-5r_2+1,2r_2-2)$ be the points on the given lines so that PQ is the line of shortest distance between the gives lines. Now direction ratios of PQ are
$2r_1-3r_2-2,-3r_1+5r_2-1,r_1-2r_2+2$
since it is perpendicular to the given lines
$2(2r_1-3r_2-2)-3(3r_1+5r_2-1)+(r_1-2r_2+2)=0$
and $3(2r_1-3r_2-2)-5(3r_1+5r_2-1)+2(r_1-2r_2+2)=0$
$\Rightarrow 14r_1-23r_2+1=0,23r_1-38r_2+3=0$
$\Rightarrow r_1=\frac{31}{3},r_2=\frac{19}{3}$
$\Rightarrow Pis(\frac{62}{3},-31,\frac{31}{3})$ and Q is $(21,-\frac{92}{3},\frac{32}{3})$ and direction ratios of PQ are $\frac{1}{3},\frac{1}{3},\frac{1}{3}$.
Hence equations of PQ as it passes through Q are
$\frac{x-21}{\frac{1}{3}}=\frac{y+\frac{92}{3}}{\frac{1}{3}}=\frac{z-\frac{32}{3}}{\frac{1}{3}}$
$\Rightarrow 3(x-21)=3y+92=3a-32$