15     Find the shortest distance between the lines 

\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Answers (1)
D Divya Prakash Singh

We have given two lines: 

\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Calculating the shortest distance between the two lines,

\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} and \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}

by the formula 

d = \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}}

Now, comparing the given equations, we obtain

x_{1} = -1,\ y_{1} =-1,\ z_{1} =-1

a_{1} = 7,\ b_{1} =-6,\ c_{1} =1

x_{2} = 3,\ y_{2} =5,\ z_{2} =7

a_{2} = 1,\ b_{2} =-2,\ c_{2} =1

Then calculating determinant 

\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} = \begin{vmatrix} 4 &6 &8 \\ 7& -6& 1\\ 1& -2& 1 \end{vmatrix} 

= 4(-6+2)-6(7-1)+8(-14+6)

= -16-36-64

=-116

Now calculating the denominator,

\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} = \sqrt{(-6+2)^2+(1+7)^2+(-14+6)^2}= \sqrt{16+36+64}

= \sqrt{116} = 2\sqrt{29}

So, we will substitute all the values in the formula above to obtain,

d = \frac{-116}{2\sqrt{29}} = \frac{-58}{\sqrt{29}} = \frac{-2\times29}{\sqrt{29}} = -2\sqrt{29}

Since distance is always non-negative, the distance between the given lines is 

2\sqrt{29} units.

 

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