# Homework Help: Mutual positions with varying paramenter

1. Oct 16, 2016

### Kernul

1. The problem statement, all variables and given/known data
In the Euclidean space $E^3$ we have the following lines:
$r : \begin{cases} 10bx + 4y - (b + 2) = 0 \\ bx + z = 0 \end{cases}$
$s : \begin{cases} 4y - 10z + 1 = 0 \\ x + 3y - 7z + 1= 0 \end{cases}$
Find the mutual position at the varying parameter $b$.

2. Relevant equations

3. The attempt at a solution
I tried both methods, with the matrix and the directional vectors and in both I find myself wrong I guess. (sorry for the long post but I really want to understand this)
Matrix method:
I have two matrices. One made of the full equations:
$(A|d) = \begin{pmatrix} 10b & 4 & 0 & -b - 2 \\ b & 0 & 1 & 0 \\ 0 & 4 & -10 & 1 \\ 1 & 3 & -7 & 1 \end{pmatrix}$
and the other one made of the same full equations minus the last column:
$A = \begin{pmatrix} 10b & 4 & 0 \\ b & 0 & 1 \\ 0 & 4 & -10 \\ 1 & 3 & -7 \end{pmatrix}$
Now:
If the rank of the first is $4$ and the second is $3$, then the two lines are askew.
If the rank of the first is $3$ and the second is $3$, then the two lines intersect in one point.
If the rank of the first is $3$ and the second is $2$, then the two lines are parallel.
If the rank of the first is $2$ and the second is $2$, then the two lines are equivalent.
To know the rank of $A$ I'll take one of the squared submatrix of it and check the determinant. If the determinant is $0$ then I'll take another squared submatrix with one less row and one lest column.
Taking the following submatrix
$A' = \begin{vmatrix} 10b & 4 & 0 \\ b & 0 & 1 \\ 0 & 4 & -10 \end{vmatrix}$
I have a determinant equal to $0$, no matter what $b$ is. So I go deeper and check this other sub matrix:
$A' = \begin{vmatrix} 10b & 4 b & 0 \end{vmatrix}$
Which has a determinant equal to $-4b$. It is different from $0$ only if $b$ is different from $0$. So, we have that the rank of it is $2$ for $b \neq 0$. This means that the other matrix must be either $3$ or $2$.
I'm going to write all the passages for the first matrix:
$(A|d) = \begin{vmatrix} 10b & 4 & 0 & -b - 2 \\ b & 0 & 1 & 0 \\ 0 & 4 & -10 & 1 \\ 1 & 3 & -7 & 1 \end{vmatrix} = -b\begin{vmatrix} 4 & 0 & -b - 2 \\ 4 & -10 & 1 \\ 3 & -7 & 1 \end{vmatrix} - \begin{vmatrix} 10b & 4 & -b - 2 \\ 0 & 4 & 1 \\ 1 & 3 & 1 \end{vmatrix} = -b [-10(4 + 3b + 6) + 7(4 + 4b + 8)] - [4(10b + b + 2) - (30b - 4)] = -b [-100 - 30b + 84 + 28b] - [48b + 8 - 30b + 4] = -16b - 2b^2 + 18b + 12 = -2b^2 - 34b + 12$
The thing is that this should be $0$, right? Why isn't it? What did I do wrong?
Directional vectors method:
I brought the two lines into parametric form, having then:
$r : \begin{cases} x = -\frac{t}{b} \\ y = 10t + b + 2 \\ z = t \end{cases}$
$s : \begin{cases} x = -\frac{1}{2}t + 2 \\ y = \frac{5}{2} - 1 \\ z = t \end{cases}$
So I have $P_r (0, b + 2, 0), \vec v_r = (-\frac{1}{b}, 10, 1), P_s (2, -1, 0), \vec v_s = (-\frac{1}{2}, \frac{5}{2}, 1)$
Putting all these in the following matrix and finding the determinant:
$\begin{vmatrix} 0 - 2 & b + 2 - (-1) & 0 - 0 \\ -\frac{1}{b} & 10 & 1 \\ -\frac{1}{2} & \frac{5}{2} & 1 \end{vmatrix} = \begin{vmatrix} - 2 & b + 3 & 0 \\ -\frac{1}{b} & 10 & 1 \\ -\frac{1}{2} & \frac{5}{2} & 1 \end{vmatrix} = +5 - \frac{b}{2} - \frac{3}{2} - 20 + 1 + \frac{3}{b} = -\frac{31}{2} - \frac{b}{2} + \frac{3}{b} = -\frac{31 + b}{2} + \frac{3}{b}$
Again, I find myself with something that should actually be $0$ independently of $b$.
What am I doing wrong?

2. Oct 16, 2016

### Staff: Mentor

Concerning the matrix: you ignored the last row in the determination of the rank, that does not work.
I don't understand the steps that follow there, but the result is not the right determinant.

3. Oct 16, 2016

### Kernul

It's not that I ignored it, it's that since I already found a determinant that is already $0$, it's useless to calculate another squared submatrix containing the other missing row. So I directly went for the smaller submatrix of 2x2.

I took the second row and did the cofactors with $b$ and $1$ (only these two since the other two numbers on the row are $0$). Since they aren't on even positions, they are taken negative, so $-b$ and $-1$. Same thing when the these two are multiplied to the other two determinants of the two matrices.

EDIT: I just noticed a mistake in the signs. My bad. It's like on WolframAlpha.

4. Oct 16, 2016

### Ray Vickson

I don't understand the question. What is meant by "mutual position"?

5. Oct 16, 2016

### Kernul

Meaning if they the two lines are parallel, intersecting or askew.

6. Oct 16, 2016

### Staff: Mentor

That only works if you start with square matrices and test possible all smaller matrices until you find one that has non-zero determinant, or until you tested all.

7. Oct 16, 2016

### Kernul

So I have to test all the small square matrices of $A$ and find a non-zero determinant? Only if I don't find any I will try with the smaller ones?

8. Oct 16, 2016

### Staff: Mentor

If you find a 3x3 matrix in A with a non-zero determinant then A has rank 3. If all 3x3 matrices in A do not have rank 3 then A cannot have rank 3 and you have to dig deeper.
It is usually easier to perform row/column operations that do not change the rank until the rank becomes obvious.

9. Oct 20, 2016

### Kernul

Oh okay, thank you.
The thing is that it feel easier for me to do it with the determinants than try randomly with row/column operations.

P. S. Sorry for the late reply.

10. Oct 20, 2016

### Staff: Mentor

It is not random, it would be like the Gauss algorithm.