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Find the shortest distance between the lines: `(x+1)/7=(y+1)/(-6)=(z+1)/1a n d(x-3)/1=(y-5)/(-2)=(z-7)/1` .

Text Solution

Verified by Experts

As shortest distance is given by
`d = (abs[[x_2-x_(1),y_(2)-y_(1),z_(2)-z_(1)],[a_(1),b_(1),c_(1)],[a_(2), b_(2),c_(2)]])/(sqrt((b_(1)c_(2)-c_(1)b_(2))^(2)+(c_(1)a_(2)-a_(1)c_(2))^(2)+(a_(1)b_(2)-a_(2)b_(1))^(2)))`
here `(a_(1),b_(1),c_(1))-= (7, -6, 1), (a_(2), b_(2),c_(2))-=(1,-2,1)`
`(x_(2)-x_(1),y_(2)-y_(1), z_(2)-z_(1))-=(4,6,8)`
so `d= abs((-16-36-64)/sqrt(16+36+64))`
`d= 116/sqrt(116)=sqrt(116)`
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