Determine whether the lines L1 and L2 are parallel, skew, or intersecting. If th

Zoe Oneal 2021-08-20 Answered
Determine whether the lines L1 and L2 are parallel, skew, or intersecting. If they intersect, find the point of intersection.
\(\displaystyle{L}{1}:{\frac{{{x}}}{{{1}}}}={\frac{{{y}-{1}}}{{-{1}}}}={\frac{{{z}-{2}}}{{{3}}}}\)
\(\displaystyle{L}{2}:{\frac{{{x}-{2}}}{{{2}}}}={\frac{{{y}-{3}}}{{-{2}}}}={\frac{{{z}}}{{{7}}}}\)

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Expert Answer

odgovoreh
Answered 2021-08-21 Author has 23828 answers
The direction vectors are the numbers in the denominators of the symmetric equations.
\(\displaystyle{L}_{{1}}:{<}{1},-{1},{3}{>}\)
\(\displaystyle{L}_{{2}}:{<}{2},-{2},{7}{>}\)
One vector is not a multiple of the other. 2 times 1 is 2, but 2 times 3 isn't 7. The lines are not parallel.
Parametric form of \(\displaystyle{L}_{{1}}\)
\(\displaystyle{\frac{{{x}}}{{{1}}}}={t}\Rightarrow{x}={t}\)
\(\displaystyle{\frac{{{y}-{1}}}{{-{1}}}}={t}\Rightarrow{y}={1}-{t}\)
\(\displaystyle{\frac{{{z}-{2}}}{{{3}}}}={t}\Rightarrow{z}={2}+{3}{t}\)
Parametric form of \(\displaystyle{L}_{{2}}\)
\(\displaystyle{\frac{{{x}-{2}}}{{{2}}}}={s}\Rightarrow{x}={2}+{2}{s}\)
\(\displaystyle{\frac{{{y}-{3}}}{{-{2}}}}={s}\Rightarrow{y}={3}-{2}{s}\)
\(\displaystyle{\frac{{{z}}}{{{7}}}}={s}\Rightarrow{z}={7}{s}\)
See if the lines intersect by setting the component parts equal to each other and seeing if there is a solution.
\(\displaystyle{t}={2}+{2}{s}\)
\(\displaystyle{1}-{t}={3}-{2}{s}\)
\(\displaystyle{2}+{3}{t}={7}{s}\)
Substitute for t
\(\displaystyle{1}-{\left({2}+{2}{s}\right)}={3}-{2}{s}\)
\(\displaystyle{1}-{2}-{2}{s}={3}-{2}{s}\)
\(\displaystyle-{1}={3}\)
This is a contradiction, so there is not solution. The lines do not intersect so they are skew.
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