Determine the lines equation in parametric form

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Randyyy

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A line [MATH]m_1[/MATH] goes through the point [MATH](1,2,2)[/MATH] and is parallel to the plane x+3y+z=1 and intersects the line [MATH]m_2 : (x,y,z)=(1+t,2-2t,1+t)[/MATH].

express the equation for [MATH]m_1[/MATH] in parametric form.

Hey,
From what I know is that I need 2 points on the line [MATH]m_1[/MATH]. I know it intersects [MATH]m_2[/MATH] so If I take t=0 I get another point which is (1,2,1). Let [MATH]P_0=(1,2,2), Q=(1,2,1) \iff \vec{u}=\vec{P_0Q}=(1,2,2)-(1,2,1)=(0,0,1)[/MATH]. Now I know that [MATH]\vec{P_0Q}=t\vec{u}[/MATH]Is this correct so far? It feels like the values for my vector [MATH]\vec{u}[/MATH] are a bit odd and somehow It feels like what I did is just nonsense.
I know that it is parallel to the plane x+3y+z=1 so maybe I don´t really need to look for a vector [MATH]\vec{u}[/MATH] but instead I should use the fact that I have the equation for the plane?
 
Randyyy said:
A line [MATH]m_1[/MATH] goes through the point [MATH](1,2,2)[/MATH] and is parallel to the plane x+3y+z=1 and intersects the line [MATH]m_2 : (x,y,z)=(1+t,2-2t,1+t)[/MATH].

express the equation for [MATH]m_1[/MATH] in parametric form.

Hey,
From what I know is that I need 2 points on the line [MATH]m_1[/MATH]. I know it intersects [MATH]m_2[/MATH] so If I take t=0 I get another point which is (1,2,1). Let [MATH]P_0=(1,2,2), Q=(1,2,1) \iff \vec{u}=\vec{P_0Q}=(1,2,2)-(1,2,1)=(0,0,1)[/MATH]. Now I know that [MATH]\vec{P_0Q}=t\vec{u}[/MATH]Is this correct so far? It feels like the values for my vector [MATH]\vec{u}[/MATH] are a bit odd and somehow It feels like what I did is just nonsense.
I know that it is parallel to the plane x+3y+z=1 so maybe I don´t really need to look for a vector [MATH]\vec{u}[/MATH] but instead I should use the fact that I have the equation for the plane?
Click to expand...

They don't tell you which point on that line it intersects, so you can't just arbitrarily pick the easiest one!

One approach might be to determine a plane in which a second point must lie, and find where the given line intersects that plane.
 
hmmm, let's give that a try.

if I rewrite my plane as x+3y+z-1=0 I know that if m1 is going to be parallel I can write it as x+3y+z+c=0. If I plug in my coordinate that I know m1 passes through I can calculate c to be -8 and create the new plane x+3y+z-8=0 and now I just pick any point that is on that plane, suppose I then let x and y = 0 and i get the solution or coordinate rather: (0,0,8) and then I construct my vector [MATH]m_1[/MATH] as follows, [MATH]\vec{m_1}=(0-1,0-2,8-2)=(-1,-2,6)[/MATH]. this gives me that [MATH](x,y,z)=t(-1,-2,6)+(1,2,2)[/MATH]Is this what you meant or did I just completely misinterpret you?
 
Randyyy said:
A line [MATH]m_1[/MATH] goes through the point [MATH](1,2,2)[/MATH] and is parallel to the plane x+3y+z=1 and intersects the line [MATH]m_2 : (x,y,z)=(1+t,2-2t,1+t)[/MATH]. express the equation for [MATH]m_1[/MATH] in parametric form.
Click to expand...
The normal to the \(x+3y+z=1\) is \(\vec{N}\). If the line in question contains \((1,2,2)~\&~(1+t,2-2t,1+t)\)
has direction vector \(\vec{D}=\left<t,-2t,-1+2t\right>\). To find \(t\), we know that \(\vec{D}\cdot\vec{N}=0\), \(t=\frac{-1}{4}\).
So the direction vector can be written \(\vec{D}=\left<1,-2,5\right>\) and the line
\(f(t) = \left\{ \begin{gathered} x = 1 + t \hfill \\ y = 2 - 2t \hfill \\ z = 2 + 5t \hfill \\ \end{gathered} \right.\)
 
Randyyy said:
if I rewrite my plane as x+3y+z-1=0 I know that if m1 is going to be parallel I can write it as x+3y+z+c=0. If I plug in my coordinate that I know m1 passes through I can calculate c to be -8 and create the new plane x+3y+z-8=0 and now I just pick any point that is on that plane, suppose I then let x and y = 0 and i get the solution or coordinate rather: (0,0,8) and then I construct my vector [MATH]m_1[/MATH] as follows, [MATH]\vec{m_1}=(0-1,0-2,8-2)=(-1,-2,6)[/MATH]. this gives me that [MATH](x,y,z)=t(-1,-2,6)+(1,2,2)[/MATH]Is this what you meant or did I just completely misinterpret you?
Click to expand...
Is that line parallel to the plane \(x+3y+z=1\)
 
Randyyy said:
hmmm, let's give that a try.

if I rewrite my plane as x+3y+z-1=0 I know that if m1 is going to be parallel I can write it as x+3y+z+c=0. If I plug in my coordinate that I know m1 passes through I can calculate c to be -8 and create the new plane x+3y+z-8=0 and now I just pick any point that is on that plane, suppose I then let x and y = 0 and i get the solution or coordinate rather: (0,0,8) and then I construct my vector [MATH]m_1[/MATH] as follows, [MATH]\vec{m_1}=(0-1,0-2,8-2)=(-1,-2,6)[/MATH]. this gives me that [MATH](x,y,z)=t(-1,-2,6)+(1,2,2)[/MATH]Is this what you meant or did I just completely misinterpret you?
Click to expand...
Check that -8. Plugging in (1, 2, 2), I get x+3y+z-8 = 1+3*2+2-8= 1, not 0.

Why do you again think you can pick a point arbitrarily? For some purposes you can, but not here. Your second point has to be on BOTH the given plane AND the given line! You've ignored the line entirely.

pka is suggesting a different approach (one I had also considered, but chose to stick more to equations than vectors). If it happens to look better to you, use it!
 
Dr.Peterson said:
Check that -8. Plugging in (1, 2, 2), I get x+3y+z-8 = 1+3*2+2-8= 1, not 0.

Why do you again think you can pick a point arbitrarily? For some purposes you can, but not here. Your second point has to be on BOTH the given plane AND the given line! You've ignored the line entirely.

pka is suggesting a different approach (one I had also considered, but chose to stick more to equations than vectors). If it happens to look better to you, use it!
Click to expand...
Yes, you are right, I am not too sure what I am doing myself, I find linear algebra a bit tricky to understand what it is I am supposed to do as opposed to regular calc.

I also noticed that I had a little error in the question, the line [MATH]m_2 : (x,y,z)=(1+t,2-2t,-1+t)[/MATH]Trying to replicate pka I get that the normal [MATH]\vec{N}=(1,3,1)[/MATH] and the direction vector [MATH]\vec{D}=(1+t-1,2-2t-2,-1+t-2)=(t,-2t,t-3)[/MATH] and we know that [MATH]\vec{N}.\vec{D}=0 \iff t=\dfrac{-3}{4}[/MATH], that yields that [MATH]\vec{D}=(-3,6,-15)[/MATH] (I multiplied [MATH]\vec{D}[/MATH] by four to get rid of the fraction.)
[MATH]x = 1-3t[/MATH][MATH]y=2+6t[/MATH][MATH]z=2-15t[/MATH]
But I notice that my answer differs dramatically from pka, did I mess it up again?
 
[QUOTE="Randyyy, post: 526922, member: 80656"
But I notice that my answer differs dramatically from pka, did I mess it up again?[/QUOTE]
Your \((t,-2t,t-3)\) should be \((t,-2t,-1+t)\)
 
Randyyy said:
I also noticed that I had a little error in the question, the line [MATH]m_2 : (x,y,z)=(1+t,2-2t,-1+t)[/MATH]Trying to replicate pka I get that the normal [MATH]\vec{N}=(1,3,1)[/MATH] and the direction vector [MATH]\vec{D}=(1+t-1,2-2t-2,-1+t-2)=(t,-2t,t-3)[/MATH] and we know that [MATH]\vec{N}.\vec{D}=0 \iff t=\dfrac{-3}{4}[/MATH], that yields that [MATH]\vec{D}=(-3,6,-15)[/MATH] (I multiplied [MATH]\vec{D}[/MATH] by four to get rid of the fraction.)
[MATH]x = 1-3t[/MATH][MATH]y=2+6t[/MATH][MATH]z=2-15t[/MATH]
But I notice that my answer differs dramatically from pka, did I mess it up again?
Click to expand...
Of course your answer is different, since you changed the problem!
 
pka said:
[QUOTE="Randyyy, post: 526922, member: 80656"
But I notice that my answer differs dramatically from pka, did I mess it up again?
Click to expand...
Your \((t,-2t,t-3)\) should be \((t,-2t,-1+t)\)
[/QUOTE]

I had noticed that I had written originally that [MATH]m_2: (x,y,z) = (1+t,2-2t,1+t)[/MATH] but it was actually incorrect, my bad, I should be more careful when I post to make sure that what I have written is correct. It should be [MATH]m_2: (x,y,z) = (1+t,2-2t,t-1)[/MATH].

So is it correct to say then that after the discovery that I had originally expressed [MATH]m_2[/MATH] incorrectly that I indeed should get that [MATH]m_1: (x,y,z)=(1-3t, 2+6t, 2-15t)[/MATH]?
 
Dr.Peterson said:
Of course your answer is different, since you changed the problem!
Click to expand...
That is true, but I expected our answers to be a bit more similar. That slight change made a big difference, bigger than I anticipated once I actually discovered it. That is stupid on my part, I should make sure that I write the problem correctly next time.
 
Randyyy said:
Your \((t,-2t,t-3)\) should be \((t,-2t,-1+t)\)
Click to expand...

I had noticed that I had written originally that [MATH]m_2: (x,y,z) = (1+t,2-2t,1+t)[/MATH] but it was actually incorrect, my bad, I should be more careful when I post to make sure that what I have written is correct. It should be [MATH]m_2: (x,y,z) = (1+t,2-2t,t-1)[/MATH].

So is it correct to say then that after the discovery that I had originally expressed [MATH]m_2[/MATH] incorrectly that I indeed should get that [MATH]m_1: (x,y,z)=(1-3t, 2+6t, 2-15t)[/MATH]?
[/QUOTE]
Yes, I get the same answer my way.
 
Thank you Dr.Peterson and pka for the help, I appreciate it greatly!
 
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