# Line parallel to plane with real parameters

1. Sep 27, 2016

### Kernul

1. The problem statement, all variables and given/known data
Find for what real parameters the line $r$ is parallel to the plane $\pi$.
$r: \begin{cases} x - 3y + 3 = 0 \\ 2y + z - 5 = 0 \end{cases}$
$\pi : 6x - (a - 1)y + 3az - 11 = 0$

2. Relevant equations

3. The attempt at a solution
So, the only method I know is to put the three equations in a matrix, find the determinant, and see for what $a$ the determinant becomes $0$.
$\begin{vmatrix} 6 & -(a - 1) & 3a & -11 \\ 1 & -3 & 0 & 3 \\ 0 & 2 & 1 & -5 \end{vmatrix}$
And I end up with $a = \frac{5}{4}$.
But, since I'm not sure about my method, does anybody know how to proceed in these cases where the exercise involves real parameters?

2. Sep 27, 2016

### Orodruin

Staff Emeritus
Determinants are only defined for square matrices. It is therefore unclear what you are doing.

Hint: Do the constants in the equations play any role in determining the directions?

3. Sep 27, 2016

### Kernul

Sorry, I didn't explain all.
I'm actually trying to find out the rank. I'll find a square submatrix and do the determinant. If the determinant is different than $0$ then the matrix has the rank equal to the number of rows of the submatrix. If the determinant is equal to $0$ then the rank is smaller than the number of rows of the submatrix. In this case I go and take a submatrix that has one row and one column less than the previous one and to the same thing as before.

4. Sep 29, 2016

### Kernul

Yes, I already have the directions of both the line and the plane.
$\vec v_r = (- 3, - 1, 2)$
$\vec n_{\pi} = (6, - a + 1, 3a)$
But I don't know what I could do with these two.

5. Sep 29, 2016

### Orodruin

Staff Emeritus
So what must be true for these vectors in order for the line to be parallel to the plane?

6. Sep 29, 2016

### Kernul

They need to have the same direction. They must be linearly dependent?

7. Sep 29, 2016

### Orodruin

Staff Emeritus
Is the plane normal vector parallel to the plane?

8. Sep 29, 2016

### Kernul

No, it's orthogonal to it.

9. Sep 29, 2016

### Orodruin

Staff Emeritus
And therefore a line is parallel to the plane if ...

10. Sep 29, 2016

### Kernul

If it's orthogonal to it. Does that mean I have to find out the angle between the two and see if the $cos\theta = 0$?

11. Sep 29, 2016

### Orodruin

Staff Emeritus
Yes, but you do not need to find the angle. You just have to figure out if the vectors are orthogonal (i.e., if the angle is 90 degrees or not). You can do this by ...

12. Sep 29, 2016

### Kernul

Scalar product? $\vec v_r * \vec n_{\pi} = v_r n_{\pi} cos\theta$
Because if they are orthogonal, the scalar product should be $0$.

13. Sep 29, 2016

### Orodruin

Staff Emeritus
Right

14. Sep 29, 2016

### Kernul

I have $6(-3) + (-1)(-a + 1) + 3a(2) = -18 + a - 1 + 6a = -19 + 7a = 0$ cioè $a = \frac{19}{7}$. Correct?