Question

Find the equation of the plane that contains the line (x=3+2t, y=t, z=8-t) and is parallel to the plane 2x+4y+8x=17. I know that if the plane is parallel to the one mentioned then it would be a multple of it but I don't really know where to go with this

Answer 1

Maybe I am confused by this problem, but with 'contains' the line they mean that \(g \in \Delta\) for one possible value, or including the entire line?

Answer 2

where \(g\) is the line and \(\Delta\) represents the plane.

Answer 3

The plain we are looking for contains the entire line

Answer 4

Ok what I see so far is the following:
\[ \Large g: \vec{r_x}= \left(\begin{matrix}3 \\ 0 \\ 8\end{matrix}\right) + \lambda \left(\begin{matrix}2 \\ 1 \\ -1\end{matrix}\right)\] So given by the dot product they are both linearly depending.

Answer 5

That is the vector equation for the line but I am not sure what you mean by linearly depending

Answer 6

\[\Large \Delta'=2x+4y+8z+c=0 \\ g \in \Delta \longrightarrow P(3/0/8) \in \Delta \]

Answer 7

I don't follow

Answer 8

It means that
\[ \large \vec{n}\cdot \vec{v}=0 \] Where n is the normal vector given by the plane, v is the direction vector given by the line.

Answer 9

What I am trying to do is shift the plane up or down, same as with a line, two lines mx+q are parallel to each other as long as m remains constant and q varies, you can do the same with the plane, by adding a constant.

Answer 10

I tried to show, that the line and plane are parallel to each other, therefore if you shift the plane to one point that is on the line, given by the vector equation, it includes all the points and therefore the entire line.

Answer 11

okay i see what you are saying there

Answer 12

Alright so we try to figure out that constant now.

Answer 13

I endup with \[ \Large \Delta'= 2x+4y+8z-70=0 \\ \Large \Delta'=2x+4x+8z=70\]

Answer 14

I am still unsure of how you did that

Answer 15

Does it seem logical to you that these two planes are parallel to each other?

\[2x+4y+8z=17 \\2x+4y+8z=12387193 \]

Answer 16

It does. They are just shifted from one another. I just don't see how to find the plane that contains the line specified

Answer 17

Alright, so you're pretty close from there, I did show above that the dot product between the normal vector given by the plane (the one that is perpendicular to the plane itself) and the direction vector of the line is zero. That means that the the normal vector and the direction vector are 90° relative to one another, therefore - the line and the plane are parallel to each other.

Now, all you got to do is find one point on the line and then shifts your plane so it includes that point, you can do that by using the same plane (because it has to be parallel to the original one given in the problem set) and just add another constant to it, the constant will then dictate if it includes that point or not.

Answer 18

If I have the point (3,0,8) on the line how do I use that to shift my plane

Answer 19

you insert it to the equation:

\[\Large \Delta':2x+4y+8z+c=0 \]
The c is responsible for the position of the plane in \(\mathbb{R^3}\) so now:
\[\Large P(3/0/8) \in \Delta' \]
means that you plug this point into the equation:

\[\Large 2\cdot (3) + 4 \cdot (0) + 8 \cdot (8)+c=0 \]

Answer 20

ohhh

Answer 21

I got c=-70

Answer 22

it is basically the same idea as with a line, if you want to find a line that is parallel to \[\Large y=2x+4 \]
but goes through the point (8/9) you just match it to the following equation
\[ \Large y=2x+q \]

Answer 23

yes, same here

Answer 24

Sorry I am having difficulty wrapping my head around all this

Answer 25

Think you could help me through a few more similar problems?

Answer 26

you can verify that answer for yourself, choose two points from the line and see if they match the equation of the plane, if that statement holds, all line are on it.

And don't you worry, it only takes a few examples. Not too easy to understand the problem in three dimensional space.

Answer 27

I sure can give it a try.

Answer 28

The plane passes through the point -1,2,1) and contains the line with symmetric equations x=2y=3x... So I know all their direction numbers are one but from there I am stuck

Answer 29

Hmm, \[x=2y=3x? \]

Answer 30

sorry 3z

Answer 31

would the parametrics be \[x=t, y=t/2,z=t/3\]

Answer 32

That's what I got myself yes:
\[\Large \vec{r_x}=t\left(\begin{matrix}1 \\ 1/2 \\ 1/3\end{matrix}\right) = t'\left(\begin{matrix}6 \\ 3 \\ 2\end{matrix}\right)\]

Answer 33

what does t prime represent or did you just scale it to be easier to work with?

Answer 34

I just scaled it.

Answer 35

so the equation of the line is\[r _{0}(1,-1,1)+t(1,1/2,1/3)\]?

Answer 36

Well it doesn't say that the line passes through the point (1,-1,1) the plane however does. The line represented above in the parametric equation actually goes through the origin, if we did that correct - all we got to do, use the vector given by the parametric equation as a normal vector of the plane and include the point:

\[\Large 6x+3y+2z+c=0 \]

Answer 37

or no i'm sorry it should be \[(0,0,0)+t(1,1/2,1/3)\] and the point on the plane would (1,-1,1)

Answer 38

shouldn't it contain bothe the line and the point?

Answer 39

hmm let me see, I think I just confused myself now with this problem, but your equation is right, mine seems dodgy though.

Answer 40

Good, well this way we can still try the vector product

Answer 41

Sorry when interrupting, I have a question to Spacelimbus. What is the purpose of the process? to find out the equation of the plane?