Is finding the intersection of two lines in R4 using the same method as in R3?

[theRukus]

To find the intersection of two lines in R3, you set the lines equal, right?
[a,b,c] + d[e,f,g] = [h,i,j] + k[l,m,n]
Then split these into three equations,
1. a + d(e) = h + k(l)
2. b + d(f) = i + k(m)
3. c + d(g) = j + k(n)

And solve for k and d, correct?

If k and d are consistent, then these values are used to find the point of intersection.

My question is: Is this same method used in R4, with two lines of 4 dimensions?

 

[arkajad]

Probably you could get some help after explaining your notation...

 

[HallsofIvy]

Yes, it is exactly the same. You would have one line given by, say,
\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}= \begin{pmatrix}a_1t+ b_1 \\ a_2t+ b_2 \\ a_3t+ b_3 \\ a_4t+ b_4\end{pmatrix}
and the other by
\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}= \begin{pmatrix}c_1s+ d_1 \\ c_2s+ d_2 \\ c_3s+ d_3 \\ c_4s+ d_4\end{pmatrix}
then you would set them equal getting 4 equations in the two unknown values s and t:
a_1t+ b_1= c_1s+ d_1
a_2t+ b_2= c_2s+ d_2
a_3t+ b_3= c_3s+ d_3
a_4t+ b_4= c_4s+ d_4

Of course, it would be unlikely for 2 numbers to satisfy all 4 equations!

In the plane, the "typical" behavior is for lines to intersect- the only non-intersecting lines are parallel lines, a very unusual situation. And, in two dimensions, we are solving two equations in two unknowns- there would not be a unique solution only if the determinant of the coefficient matrix were 0.

In three dimensions, the "typical" behavior is for lines to be skew- not intersecting. In order to be able to solve three equations for two unknown values, the equations must NOT be independent.

And the situation is worse in four dimensions. Two lines intersecting is a very special situation indeed.