+0  
 
0
76
4
avatar+777 

For a i got r= (-2,3,4) + λ(3,2,-7)

for b so far i wanted to identify a point A on l1 and a point b on l2 

i assumed the two lines are parallel as they have the same direction vector 

\(A=\begin{pmatrix} -2+3λ\\ 3+2λ\\ 4-7λ \end{pmatrix}\) \(B= \begin{pmatrix} 3\\ 1+2λ\\ -2-7λ \end{pmatrix}\)

used this to get 

\(AB= \begin{pmatrix} 5-3λ\\ -2\\ -6 \end{pmatrix}\)

used this and my direction vector to find λ=53/9 via dot product

substituted and calculated the magnitude which gave me 14.26

but my answer should be 3.57

please identify what went wrong, thanks

 Feb 10, 2019
 #1
avatar
+2

The equation of the first line can be written as r1 = (-2, 3, 4) + s(3, 2, -7), (as you calculated),

and the equation of the second as r2 = (0, 1, -2) + t(3, 2, -7), s and t arbitrary parameters.

A vector d connecting two points, one on each line, would be given by

d = r2 - r1 = (2, -2, -6) + (t - s)(3, 2 ,-7) = (2, -2, -6) + w(3, 2, -7), w for convenience.

The distance between the two points will be the magnitude of this vector, so what you are looking for is the minimum value of

sqrt{(2 + 3w)^2 + (-2 + 2w)^2 + (-6 -7w)^2} = sqrt{62w^2 + 88w + 44} and that turns out to be 3.5741 (4dp).

 

Tiggsy

 Feb 10, 2019
 #4
avatar+777 
0

thanks!

YEEEEEET  Feb 10, 2019
 #2
avatar+10069 
+2

Shortest disdance between two lines via vektor equation.

laugh

 Feb 10, 2019
edited by Omi67  Feb 10, 2019
 #3
avatar+777 
0

ah ok i realised where ive gone wrong thanks a lot 

YEEEEEET  Feb 10, 2019

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