# Thread: Vectors - Scalar equation of a plane/symmetric equation of a plane problem

1. ## Vectors - Scalar equation of a plane/symmetric equation of a plane problem

Hey, thanks for coming into my thread!

I have a problem involving vectors that I do not know how to start solving.

Question
Find the value of k for which the plane $\displaystyle kx + 4y + 2z - 6 = 0$ is parallel to the line $\displaystyle \frac{x - 3}{5} = \frac{y}{1} = \frac {z}{-3}$.

Solution
I know that (k, 4, 2) will be perpendicular to a direction vector on the plane given in the symmetric equation.
Rearranging the symmetric equation gives me:
$\displaystyle x = 5t + 3$
$\displaystyle y = t$
$\displaystyle z = -3t$

Which I believe gives me a direction vector on that plane of: (5, 1, -3).

With this information, how can I find the value of k? If I had a second direction vector of the parallel plane I could do a cross product, so is it possible to find that?

Thanks!

2. Originally Posted by Kakariki
Hey, thanks for coming into my thread!

I have a problem involving vectors that I do not know how to start solving.

Question
Find the value of k for which the plane $\displaystyle kx + 4y + 2z - 6 = 0$ is parallel to the line $\displaystyle \frac{x - 3}{5} = \frac{y}{1} = \frac {z}{-3}$.

Solution
I know that (k, 4, 2) will be perpendicular to a direction vector on the plane given in the symmetric equation.
Rearranging the symmetric equation gives me:
$\displaystyle x = 5t + 3$
$\displaystyle y = t$
$\displaystyle z = -3t$

Which I believe gives me a direction vector on that plane of: (5, 1, -3).

With this information, how can I find the value of k? If I had a second direction vector of the parallel plane I could do a cross product, so is it possible to find that?

Thanks!
All your considerations and calculations are OK.

If the direction vector of the line is perpendicular to the normal vector of the plane then the plane must be parallel to the line.

3. Originally Posted by earboth
All your considerations and calculations are OK.

If the direction vector of the line is perpendicular to the normal vector of the plane then the plane must be parallel to the line.
Would doing a dot product help? setting it equal to zero and then solving for k?

4. Originally Posted by Kakariki
Would doing a dot product help? setting it equal to zero and then solving for k?
Correct!

5. Originally Posted by earboth
Correct!
SWEET! Thank you very much for your help.