Finding equations of two lines

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Homework Statement



Find 2 lines in R3 that are not parallel and do not intersect.

Homework Equations



orthogonal: a.b=0
parallel: a=tb

The Attempt at a Solution



(1,1,1) and (2,3,4)

(1,1,1).(2,3,4) = 7
and there does not exist a t such that
t(1,1,1) = (2,3,4)
 
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Is that all you're given? Any 2 lines would do?

Your answer doesn't include the starting point of the line vector it's supposed to be in: (x0,y0,z0) + t(a,b,c). The point (x0,y0,z0) isn't specified. That's why you haven't shown why they do not intersect.
 
If fact, your answers are not lines at all! What you give look like points but I guess you mean them as vectors- and I presume you mean them as "direction" vectors for the lines. The fact that "(1,1,1).(2,3,4) = 7" shows that they are not perpendicular (not really relevant) and the fact that "there does not exist a t such that t(1,1,1) = (2,3,4)" shows that they are not parallel. Neither of those means they do not intersect.
As defennnder said, you need to write them as lines with a parameter such as t. To do that, you will have to specify a point on them- and whether or not they intersect will depend on that point. For example, the lines \vec{r_1}= t\vec{i}+ t\vec{j}+ t\vec{k} and \vec{r_2}= 2t\vec{i}+ 3t\vec{j}+ 4t\vec{k} have direction vectors \vec{i}+ \vec{j}+ \vec{k} and 2\vec{i}+ 3\vec{j}+ 4\vec{k} respectively and intersect at (0, 0, 0). You need to find skew lines.
 
First try to get the geometrical picture clear. Can you describe in words two lines that would do?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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