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So that wasn't a question but a statement. At least I can't figure out how they intersect. Question is: Find an equation of the hyperplane that contains the lines q(t) = (t,t,t,1) and f(t) = (1,t,1+t,t), t $\in$ $\Re$

I need two things to answer the question. First, find the point. So the lines are clearly not parallel but does q(t) = f(t) ever? Not in my calculations. So they don't intersect. I'd appreciate if someone could verify for me or tell me how wrong I am. Second find 't'. Haven't done that.

Jacob
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3 Answers3

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Suppose there exists $t,s \in \mathbb{R}$ such that $q(t) = f(s)$. Then $(t,t,t,1) = (1,s,s+1,s) \implies (t-1,t-s,t-s-1,1-s) = (0,0,0,0)$. Notice that $t-1 = 0 \implies t=1$ and $1-s = 0 \implies s =1$. However, $t-s-1 = 1 -1 -1 = -1 \neq 0$. Thus, this is impossible

Kyle Gannon
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The lines don't intersect. Imagine two lines in three-dimensional space that are in different directions and don't intersect. For example, one pointing north-south, at ground level, and one pointing east-west, ten feet above ground level. They are called skew lines.
This hyperplane is a three-dimensional subset of $\mathbb{R}^4$. It is the set of points $(w,x,y,z)$ for which $aw+bx+cy+dz=e$.
You are to find the numbers $a,b,c,d$ and $e$.
$at+bt+ct+d=e$ for every $t$, so both $a+b+c+d=e$ and $0+0+0+d=e$.
You get two more equations from the other line.

Empy2
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You don't want the two lines to be parallel or to intersect.

If you were working in $\mathbb R^3$ (a three-dimensional space) and you were looking for a two-dimensional plane containing two lines, the lines would have to be parallel or else intersect, because there are no other kinds of lines contained in a two-dimensional plane.

But you have points with four coordinates, that is, you are working in $\mathbb R^4$, and you need to identify a hyperplane in that space. Presumably, you are looking for one of the many three-dimensional hyperplanes that exist in that four-dimensional space.

If your two lines were parallel or intersected, they would lie within a single two-dimensional plane within $\mathbb R^4$, and there would be infinitely many three-dimensional hyperplanes that contained that plane (and therefore contained both lines). Your only hope to identify the desired hyperplane uniquely is if the two lines do not lie in a single plane.

An analogy in three-dimensional space is that you can uniquely identify a two-dimensional plane using just three points, but only if the three points are not all on the same line.

David K
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