# [SOLVED] Determine if a line and plane are parallel

• Mar 6th 2010, 03:08 PM
thekrown
[SOLVED] Determine if a line and plane are parallel
x = -5 - 4t
y = 1 - t
z = 3 +2t

and x + 2y + 3z - 9 = 0

Now how do I approach a problem like this? I am capable of calculating dot products, cross products, finding vectors from two points.

What is involved in this? Just a little boost will help and I'll try to solve it. Thanks!
• Mar 7th 2010, 01:18 AM
mr fantastic
Quote:

Originally Posted by thekrown
x = -5 - 4t
y = 1 - t
z = 3 +2t

and x + 2y + 3z - 9 = 0

Now how do I approach a problem like this? I am capable of calculating dot products, cross products, finding vectors from two points.

What is involved in this? Just a little boost will help and I'll try to solve it. Thanks!

Consider the normal to the plane and the vector in the direction of the line and take the dot product.
• Mar 7th 2010, 11:40 AM
thekrown
This is where I have trouble. I'm used to dealing with points, I have difficulty reading the way the information is displayed. I'm not really sure which one is the plane.

Okay, I believe x + 2y + 3z - 9 = 0 is the plane and the other info can be used to find a point. How is this read? I know it is a parametric equation of the line, so it should be

x = -5
y = 1
z = 3

If I am reading this correctly, I can use the distance between point and a plane formula and we get -3 / square root 14.

Now that we have the distance, I'm supposed determine if the line and plane are parallel. I could use some help on this part.
• Mar 7th 2010, 01:27 PM
mr fantastic
Quote:

Originally Posted by thekrown
This is where I have trouble. I'm used to dealing with points, I have difficulty reading the way the information is displayed. I'm not really sure which one is the plane.

Okay, I believe x + 2y + 3z - 9 = 0 is the plane and the other info can be used to find a point. How is this read? I know it is a parametric equation of the line, so it should be

x = -5
y = 1
z = 3

If I am reading this correctly, I can use the distance between point and a plane formula and we get -3 / square root 14.

Now that we have the distance, I'm supposed determine if the line and plane are parallel. I could use some help on this part.

You need to extensively review your class notes. Below are things you are expected to know.

The cartesian equation of a plane can be written ax + by + cz = d and a normal vector to this plane is <a, b, c>.

The parametric equations of a line can be written $x = x_0 + \alpha t$, $y = y_0 + \beta t$, $z = z_0 + \gamma t$ and a vector in the direction of this line is $<\alpha, \, \beta, \, \gamma>$.

Now apply the defintion you will undoubtedly have been given on establishing whether a line and plane are parallel.
• Mar 7th 2010, 02:24 PM
thekrown
Okay, I think I got this. My teacher and textbook use different notations, or there are different notations so the examples I have in class are not exactly as those in the textbook but I think I got it.

With

x = -5 - 4t
y = 1 - t
z = 3 +2t

and x + 2y + 3z - 9 = 0

I can sub the x,y,z into the equation. Then in order to find out if it is parallel I have this note:

at + b = 0
a = 0, then b = 0

If we get 0 = 0 then any t is a solution and is parallel.

If we get b = 0 or b =/= 0 then perpendicular.

In this problem, we get

(-5-4t) + 2(1-t) + 3(3+2t) - 9 = 0
-5 -4t + 2 - 2t + 9 + 6t - 9 = 0

Since t = 0 then a = 0 and so b is 0. So by the above notes we should get 0 = 0. I actually get -3=0.

Following my notes, I would say this is parallel. I just don't understand why I get -3 = 0.
• Mar 7th 2010, 02:44 PM
mr fantastic
Quote:

Originally Posted by thekrown
Okay, I think I got this. My teacher and textbook use different notations, or there are different notations so the examples I have in class are not exactly as those in the textbook but I think I got it.

With

x = -5 - 4t
y = 1 - t
z = 3 +2t

and x + 2y + 3z - 9 = 0

I can sub the x,y,z into the equation. Then in order to find out if it is parallel I have this note:

at + b = 0
a = 0, then b = 0

If we get 0 = 0 then any t is a solution and is parallel.

If we get b = 0 or b =/= 0 then perpendicular.

In this problem, we get

(-5-4t) + 2(1-t) + 3(3+2t) - 9 = 0
-5 -4t + 2 - 2t + 9 + 6t - 9 = 0

Since t = 0 then a = 0 and so b is 0. So by the above notes we should get 0 = 0. I actually get -3=0.

Following my notes, I would say this is parallel. I just don't understand why I get -3 = 0.

I don't know what you're doing. Taking the dot product, as I suggested in my first post, the normal to the plane is obviously perpendicular to the line. It follows by defniition that the line and plane are parallel.
• Mar 7th 2010, 02:50 PM
thekrown
I don't know either. My teacher speaks with a thick Russian accent and understanding the material is hard enough in itself. Please bear with me.

Anyways, I believe I finally got it. I'm learning more on this subject with google and this website than my actual class. Thank god for this forum.

So, again, we have

x = -5 - 4t
y = 1 - t
z = 3 +2t

and x + 2y + 3z - 9 = 0

The vector of the line is (-5,1,3) and it's direction vector is (-4,-1,2)

The vector of the plane is (1,2,3).

I take the dot product, if it is 0 then they are parallel.

The dot product is -4 -2 +6 = 0 and so here we get the 0 = 0.

I'm going to have to clarify the notes with my teacher, chances are I'm just confusing two different things.