Mutual positions with varying paramenter

[Kernul]

Homework Statement

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 $b$.

Homework Equations

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?

 

[mfb]

Concerning the matrix: you ignored the last row in the determination of the rank, that does not work.

I'm going to write all the passages for the first matrix:

I don't understand the steps that follow there, but the result is not the right determinant.

 

[Kernul]

Concerning the matrix: you ignored the last row in the determination of the rank, that does not work.

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 don't understand the steps that follow there, but the result is not the right determinant.

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.

 

[Ray Vickson]

Homework Statement

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 $b$.

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

 

[Kernul]

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

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

 

[mfb]

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.

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.

 

[Kernul]

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.

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?

 

[mfb]

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.

 

[Kernul]

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.

Oh okay, thank you.

It is usually easier to perform row/column operations that do not change the rank until the rank becomes obvious.

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.

 

[mfb]

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