Question

The shortest distance between the lines $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$ and $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$ is

• A. $7\sqrt{3}$
• B. $6\sqrt{3}$
• C. $4\sqrt{3}$
• D. $5\sqrt{3}$

Answer 1

Answer: (B)

Shortest distance between two lines
$\frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{z-z_1}{a_3}\&$
$\frac{x-x_2}{b_1}=\frac{y-y_2}{b_2}=\frac{z-z_2}{b_3}$ is given as
$\frac{\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|}{\sqrt{{\left(a_1{b}_3-a_3{b}_2\right)}^2+{\left(a_1{b}_3-a_3{b}_1\right)}^2+{\left(a_1{b}_2-a_2{b}_1\right)}^2}}$
$\frac{\left|\begin{array}{ccc}5-(3)& 2-(-5)& 4-1\\ 1& 2& -3\\ 1& 4& -5\end{array}\right|}{\sqrt{(-10+12{)}^2+(-5+3{)}^2+(4-2{)}^2}}$
$\frac{\left|\begin{array}{ccc}8& 7& 3\\ 1& 2& -3\\ 1& 4& -5\end{array}\right|}{\sqrt{(2{)}^2+(2{)}^2+(2{)}^2}}$
$=\frac{|8(-10+12)-7(-5+3)+3(4-2)|}{\sqrt{4+4+4}}$
$=\frac{16+14+6}{\sqrt{12}}=\frac{36}{\sqrt{12}}=\frac{36}{2\sqrt{3}}$
$=\frac{18}{\sqrt{3}}=6\sqrt{3}$