The shortest distance between the lines x-3/1=y-5/-2=z-7/1 and x+1/7=y+1/-6=z+1/1 is

Correct Answer is : 2292{29}229 units

Correct Answer is : 2292{29}229 units

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CGPET 2013 Solved Paper

Section: Mathematics

Question : 129 of 150

Marks: +1, -0

\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}

Solution:

Given, lines are

\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

and

\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}

Shortest distance between the lines,

d = \left| \frac{\begin{vmatrix} -1-3 & -1-5 & -1-7 \\ 1 & -2 & 1 \\ 7 & -6 & 1 \end{vmatrix}}{\sqrt{(-2+6)^2+(7-1)^2+(-6+14)^2}} \right|

= \left| \frac{\begin{vmatrix} -4 & -6 & -8 \\ 1 & -2 & 1 \\ 7 & -6 & 1 \end{vmatrix}}{\sqrt{4^2+6^2+8^2}} \right|

= \left| \frac{-4(-2+6)+6(1-7)-8(-6+14)}{\sqrt{16+36+64}} \right|

= \left| \frac{-16-36-64}{\sqrt{116}} \right|

= \left| \frac{-16-36-64}{\sqrt{116}} \right|

= \left| \frac{-116}{\sqrt{116}} \right| = \sqrt{116}

= 2\sqrt{29}

units

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