Find the shortest distance between the lines given. l<sub>1</sub>:\frac{x-5}{2}=\frac{y-2}{5}=\frac{z-1}{4} l<sub>2</sub>:\frac{x+4}{3}=\frac{y-7}{6}=\frac{z-3}{7}

Category: QuestionsFind the shortest distance between the lines given. l<sub>1</sub>:\frac{x-5}{2}=\frac{y-2}{5}=\frac{z-1}{4} l<sub>2</sub>:\frac{x+4}{3}=\frac{y-7}{6}=\frac{z-3}{7}
Editor">Editor Staff asked 11 months ago

Find the shortest distance between the lines given.
 
l<sub>1</sub>:\frac{x-5}{2}=\frac{y-2}{5}=\frac{z-1}{4}
 
l<sub>2</sub>:\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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

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}}.