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Determine wether the lines L_{1}, L_{2}are parallel, skew or intersecting. If they intersect, find the point of intersection.

L_{1}= [tex]\left(\begin{array}{ccc}x\\y\\z\end{array}\right) =[/tex] [tex]\left(\begin{array}{ccc}1\\2\\0\end{array}\right)[/tex] [tex]+ t \left(\begin{array}{ccc}1\\-1\\3\end{array}\right)[/tex]

L_{2}= [tex]\left(\begin{array}{ccc}x\\y\\z\end{array}\right) =[/tex] [tex]\left(\begin{array}{ccc}2\\1\\4\end{array}\right)[/tex] [tex]+ s \left(\begin{array}{ccc}-1\\2\\1\end{array}\right)[/tex]

For some [tex]t, s \in R[/tex]

3. The attempt at a solution

I believe L_{1}& L_{2}are not parallel because their direction vectors aren't multiples of each other.

The dot product of their direction vectors are zero; this tells us that they are orthogonal/skew and not parallel.

To find the intersection point we say: L_{1}= L_{2}

x_{1}= x_{2}

[tex]\left(\begin{array}{ccc}1\\2\\0\end{array}\right)[/tex] [tex]+ t \left(\begin{array}{ccc}1\\-1\\3\end{array}\right) =[/tex] [tex]\left(\begin{array}{ccc}2\\1\\4\end{array}\right)[/tex] [tex]+ s \left(\begin{array}{ccc}-1\\2\\1\end{array}\right)[/tex]

yields the following system of linear equations;

1+t = 2-s

2-t = 1+2s

3t = 4+s

I think maybe it can be re-written as:

t+s = 1

-t-2s = -1

3t-s = 4

Is my working correct so far? If yes, can you guys show me how to solve this set of simultaneous equations to find the actual point of intersection please. Thanks.

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# Homework Help: Intersection of Lines

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