# Math Help - Show that a line is parallel to a plane

1. ## Show that a line is parallel to a plane

The question:
Show that the line x = t(2 1 3) is parallel to the plane 3x - 3y - z = 2.

I'm not sure how to check if it's parallel. I converted the plane to parametric form, but the line doesn't seem to be a scalar multiple of either directional vectors of the plane. How do I solve this?

Thanks.

2. Hello, Glitch!

Show that the line $x \,=\, t\langle2\;1\;3\rangle$ is parallel to the plane $3x - 3y - z \:=\: 2.$

The direction of the line is: . $\vec v \:=\:\langle 2,\:1,\:3\rangle$

The normal of the plane is: . $\vec n \:=\:\langle 3,\:\text{-}3,\:\text{-}1\rangle$

If the line is parallel to the plane, then: . $\vec v \:\perp\:\vec n$

Got it?

3. Thank you! I didn't know the normal was just the coefficients of the plane equation. Not sure why my textbook doesn't mention that...

4. Here's another way to do it:

A line is parallel to a plane if and only if it does not intersect that plane. So what happens if you try to find where that line intersects the plane?

The line is x= 2t, y= t, z= 3t and the plane is 3x- 3y- z= 2. Putting those expressions for x, y, and z into the equation of the plane, 3(2t)- 3(t)- (3t)= 0t= 2. Obviously, that cannot be solved: the line and plane do not intersect. Of course the left hand side is essentially the dot product of the two vectors <2, 1, 3> and <3, 3, -1>, the one in the direction of the line, the other perpendicular to the plane. Since the two vectors have 0 dot product, they are perpendicular which means that the line is parallel to the plane.

5. Originally Posted by Glitch
The question:
Show that the line x = t(2 1 3) is parallel to the plane 3x - 3y - z = 2.

I'm not sure how to check if it's parallel. I converted the plane to parametric form, but the line doesn't seem to be a scalar multiple of either directional vectors of the plane. How do I solve this?

Thanks.

if you have line $l$:

$\displaystyle \frac {x-x_1}{m} = \frac {y-y_1}{n} = \frac {z-z_1}{p}$

and plane $\alpha$ :

$Ax+By+Cz+D=0$

so you have vectors :

$\vec{q} =(m,n,p)$ and $\vec {n} = (A,B,C)$

so to line and plane to be parallel it must be that :

$(\vec{n},\vec{q}) =0$

or

$Am+Bn+Cp =0$