Question

Find the shortest distance between the lines given.

l₁:\(\frac{x-5}{2}=\frac{y-2}{5}=\frac{z-1}{4}\)

l₂:\(\frac{x+4}{3}=\frac{y-7}{6}=\frac{z-3}{7}\)

(a) \(\frac{115}{\sqrt{134}}\)

(b) \(\frac{115}{\sqrt{184}}\)

(c) \(\frac{115}{134}\)

(d) \(\frac{\sqrt{115}}{134}\)

I had been asked this question in an interview for internship.

My enquiry is from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12

Answer 1

The correct option is (a) \(\frac{115}{\sqrt{134}}\)

Best explanation: The shortest distance between two lines in cartesian form is given by:

l1:\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\)

l2:\(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)

∴d=\(\left |\frac{\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}}{\sqrt{(b_1 c_2-b_2 c_1)^2+(c_1 a_2-c_2 a_1)^2+(a_1 b_2-a_2 b_1)^2}}\right |\)

d=\(\left |\frac{\begin{vmatrix}-9&5&2\\2&5&4\\3&6&7\end{vmatrix}}{\sqrt{√(35-24)^2+(12-14)^2+(12-15)^2}}\right |\)

d=\(\left |\frac{-9(35-24)-5(14-12)+2(12-15)}{\sqrt{11^2+2^2+3^2}}\right |\)

d=\(\left |\frac{-99-10-6}{\sqrt{134}}\right |\)

d=\(\frac{115}{\sqrt{134}}\).