Question

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

• A. $\sqrt{30}$
• B. $2\sqrt{30}$
• C. $5\sqrt{30}$
• D. $3\sqrt{30}$

Answer 1

The correct option is D $3\sqrt{30}$

The lines are,

$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ ------ ( 1 )

$\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ ------ ( 2 )

Comparing the equations with,

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$

We get,

$\Rightarrow$ $x_1=3,y_1=8,z_1=3$ and $a_1=3,b_1=-1,c_1=1$

$\Rightarrow$ $x_2=-3,y_2=-7,z_2=6$ and $a_2=-3,b_2=2,c_2=4$

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$ $=\left|\begin{array}{ccc}-6& -15& 3\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|$

$=-6(-4-2)+15(12+3)+3(6-3)$

$=-6(-6)+15\left(15\right)+3\left(3\right)$

$=36+225+9$

$=270$

$\Rightarrow$ $\sqrt{(a_1{b}_2-a_2{b}_1{)}^2+(b_1{c}_2-b_2{c}_1{)}^2+(c_1{a}_2-c_2{a}_1{)}^2}$

$\Rightarrow$ $\sqrt{\left[\right(3\left)\right(2)-(-3\left)\right(-1){]}^2+[(-1)\left(4\right)-\left(2\right)\left(1\right){]}^2-\left[\right(1\left)\right(-3)-(4\left)\right(3){]}^2}$

$\Rightarrow$ $\sqrt{(6-3{)}^2+(-4-2{)}^2+(-3-12{)}^2}$

$\Rightarrow$ $\sqrt{(3{)}^2+(-6{)}^2+(-15{)}^2}$

$\Rightarrow$ $\sqrt{9+36+225}$

$\Rightarrow$ $\sqrt{270}$

Shortest distance between line is $d$.

$\Rightarrow$ $d=\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{(a_1{b}_2-a_2{b}_1{)}^2+(b_1{c}_2-b_2{c}_1{)}^2+(c_1{a}_2-c_2{a}_1{)}^2}}\right|$

$\Rightarrow$ $d=\left|\frac{270}{\sqrt{270}}\right|$

$\Rightarrow$ $d=\frac{\sqrt{270}\times \sqrt{270}}{\sqrt{270}}$

$\Rightarrow$ $d=\sqrt{270}$

$\Rightarrow$ $d=3\sqrt{30}$