Line parallel to plane with real parameters

[Kernul]

Homework Statement

Find for what real parameters the line $r$ is parallel to the plane $\pi$.
$r: \begin{cases} x - 3y + 3 = 0 \\ 2y + z - 5 = 0 \end{cases}$
$\pi : 6x - (a - 1)y + 3az - 11 = 0$

Homework Equations

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?

 

[Orodruin]

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?

 

[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.

 

[Kernul]

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

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.

 

[Orodruin]

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.

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

 

[Kernul]

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

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

 

[Orodruin]

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

Is the plane normal vector parallel to the plane?

 

[Kernul]

Is the plane normal vector parallel to the plane?

No, it's orthogonal to it.

 

[Orodruin]

No, it's orthogonal to it.

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

 

[Kernul]

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

 

[Orodruin]

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

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 ...

 

[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 $0$.

 

[Orodruin]

Right

 

[Kernul]

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