- #1

Kernul

- 211

- 7

## 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?