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

Answer: (D)

$l_1:\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$
$l_2:\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$
Shortest distance between two lines is,
$=\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{\left(b_1{c}_2-b_2{c}_1\right)}^2+{\left(c_1{a}_2-c_2{a}_1\right)}^2+{\left(a_1{b}_2-a_2{b}_1\right)}^2}}\right|$
$=\left|\frac{\left|\begin{array}{ccc}-3-3& -7-8& 6-3\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|}{\sqrt{{\left(-4-2\right)}^2+{\left(-3-12\right)}^2+{\left(6-3\right)}^2}}\right|$
$=\left|\frac{\left|\begin{array}{ccc}-6& -15& 3\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|}{\sqrt{{\left(-6\right)}^2+{\left(-15\right)}^2+{\left(3\right)}^2}}\right|$
$=\frac{3\left(6-3\right)-1\left(-12-45\right)+4\left(6+45\right)}{\sqrt{36+225+9}}=\frac{9+57+204}{\sqrt{270}}$
$=\frac{270}{\sqrt{270}}=\sqrt{270}=3\sqrt{30}$