- #1
roam
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Determine wether the lines L1, L2 are parallel, skew or intersecting. If they intersect, find the point of intersection.
L1 = [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]
L2 = [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 L1 & L2 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: L1 = L2
x1 = x2
[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.