Find parametric equation for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1-t, z=2t and intersects this line

- Jhannybean

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- schrodinger

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- anonymous

Is this calculus BC?

- anonymous

What's the difference?

- anonymous

Wow, you must be the smartest person in your highschool!

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## More answers

- anonymous

What is paramedrics mean?

- Jhannybean

@terenzreignz

- Jhannybean

HALP Terence!

- terenzreignz

Let's see what I can do...

- terenzreignz

We need a point (already have one) and a direction vector from what I can remember.
Your thoughts?

- Jhannybean

Yep. parametric equations give us the vector, \(\vec v\) and \( P_0\).

- Psymon

I think I might know how to do it, but Im not confident enough, haha.

- terenzreignz

I... don't.
Jhanny's forcing me to study... and that's just SICK D:
Teach her a lesson for me @Psymon >:D

- Jhannybean

Lmao what...

- Jhannybean

\[\large \vec v = <1,-1,2>\] i found it but how the hell is it parallel to the line L.... |:<

- anonymous

Are you all in paramedrics?

- Psymon

Well....I think we need a point on the line, too. Because if we're going to get something perpendicular to the given line, we need a normal vector. And to get a normal vector out of this we would have to cross two vectors. We have the direction vector given by the line, but we need to pick a point on that line and then make a second vector with the two points we have.

- Jhannybean

oh, and \(P_0 = (1,1,0) \therefore \vec r_0 = <1,1,0>\)

- Psymon

Yeah, when t = 0 you can say that is a point on the line, absolutely. So now use that point and the given point to make another vector.

- Jhannybean

Alright.
P= (0,1,2)
\(P_0 = (1,1,0)\)
\[\vec {P_{0}P} = <1-0,1-1,0-2>= <1,0,-2>\]

- Psymon

Right. Now cross that vector with the direction vector given by the line.

- Jhannybean

Or wait... did i do it backwards?... ?_?

- Jhannybean

But why are we trying to find a normal vector?

- Psymon

Because a normal vector is always going to be perpendicular to the vectors you're crossing.

- Jhannybean

How would the normal vector help us find the intersection...
And cross producting 2 vectors GIVES us the normal vector. So then is the normal vector perpendicular to all other vectors in the plane?

- Psymon

Yeah, a normal vector is perpendicular to all vectors in the plane.

- Jhannybean

But ok. Normal vector
|dw:1378709367099:dw|

- Jhannybean

\[\large (0-2)i - (0-2)j+(1+1)k\]\[\large -2i+2j+k\]

- Psymon

Crossed the wrong two.
So the direction vector from the given line we have correct, 1, -1, 2. But then we set t equal to 0 and found a point on the line of 1, 1, 0. Next we take the point 1, 1, 0, and 0, 1, 2, and find a component vector out of the two. Doing that givves us 1, 0, -2. This 1, 0, -2 is the other vector we need to use to cross:
|dw:1378709680738:dw|
Thats the result we should have.

- Jhannybean

Hmm..I havent learned about setting t=0 and all that good stuff. Hmmm...

- Jhannybean

What do we do from there? ..

- Psymon

Well, its a way to find a point on the line, set t = to something. But yeah, since the normal vector is perpendicular to the vectors we use, this means we know it intersects the line we have been given. So now all we do is tack on the given point, (0,1,2), to the direction normal vector we just found. So that means the line we have would be x = 2t, y = 1+ 4t, z = 2 + 2t.

- Jhannybean

so \(P=(0,1,2) = P_0=(x_0,y_0,z_0)\)?

- Psymon

Pretty much. Since we have to have that point on the line.

- Jhannybean

Hmm... I found someone that answered this same question, but it was done completely different. http://answers.yahoo.com/question/index?qid=20100628225724AABfYAn

- Psymon

Oh O.o Interesting.

- Jhannybean

How can i cross check the parameters to see if theyre correct?...

- Jhannybean

How can i cross check the parameters to see if theyre correct?...

- Psymon

Difficult to do since what I did didn't actually give you the intersection point. The method used there I haven't seen before, not even in my text. Makes sense I guess when I stare at it enough, lol.

- Jhannybean

And I wonder why theyre using.. a family of planes normal to L?

- Jhannybean

Im unable to imagine what this would look like...

- Psymon

The idea behind what he's doing is the equation of a plane a(x-xo) + b(y-yo) + c(z-z0) + d....the letters a b and c make up a vector normal to the plane given. So what he does is he invents a plane that intersects the given point.

- Jhannybean

i see.

- Psymon

|dw:1378711466764:dw|
This was his thought process. He invents the plane I drew, using the given direction vector to come up with the a, b, c components and the given point to fill in the missing info.

- Jhannybean

Ah...
Looking back at the vector \(\vec {P_0P}\) we created, i believe it was supposed to be <-1,0,2> Yeah?

- Psymon

Possibly. I usually somehow pick the wrong order to do it in :/ Unless it tells me its point P or point Q I mix them up xD But we can always do the given method and try that.

- Jhannybean

Ah... I got \(\vec n=<-2,-4,1>\)

- Psymon

x-y+2z + d = 0
(0)-(1)+2(2)+d = 0
3+d = 0, d =-3
family of planes = x-y+2z - 3 = 0
Substitute the line into the plane equation
(1+t)-(1-t) + 2(2t) - 3 = 0
1 + t -1 + t + 4t = 3
6t = 3
t = (1/2)
Substituting t into the line we get:
x = 1 +(1/2), y = 1-(1/2), z = 2(1/2)
point = (3/2, 1/2, 1)
Using this point and the old point to get a direction vector we get:
(0-3/2, 1-1/2, 2-1) = (-3/2, 1/2, 1) - direction. Direction + point gives us the line:
x = (-3/2)t, y = 1 + (1/2)t, z = 2 + t
Looks like it turns out, too. We can check to make sure this line is truly perpendicular to the given line by dotting them. If two vectors are perpendicular, their dot product is 0.
(1*(-3/2)) + (-1*(1/2)) + (2*1)
(-1.5 - .5 + 2) = 0
So yeah, looks like that's the right method. Just have never seen itbefore xD

- Jhannybean

O_o so tehres more than one solution i guess? o:

- Psymon

I think this solution gives you more information than mine does xD

- Jhannybean

Hmm... it's just not a method im familiar with. We could also take your solution and dot product that as well.

- Psymon

Yeah, mine works, I just realized I put the wrong answer despite what I think is the right method xD

- Jhannybean

Wait what? o-o

- Psymon

My z in my answer shouldve been 2 + t

- Psymon

What I mean is I put the correct normal vector of 2i + 4j + k, but mixed up the numbers when putting it into the solution.

- Jhannybean

Oh I see it.

- Psymon

But yeah, the original vector of 1, -1, 2 dotted with mine:
(1*2 + -1*4 + 2*1)
2 - 4 + 2 = 0

- Jhannybean

Lets see... how would the dot product work on this....

- Jhannybean

(sorry im a little slow at this, i just started learning how to compute these today x_x)

- Psymon

No worries xD I still got a lot to learn about these things, too x_x Its hard to know what to do, even if I know how to do all the operations involved.

- Psymon

Ive just been going through this chapter in my textbook xD

- Jhannybean

Have you already taken this class before? o:

- Psymon

Nope. Im going to take it next semester. I couldnt take it this semester because my discrete math class interfered. But I have the textbook and ive been trying to study.

- Jhannybean

Hmm.. well, when i redid the cross product, i got -2i -4j+k
\(P_0 = (0,1,2) \) and \(\vec n= <-2,-4,1>\) therefore the parametric equations are \[x=-2t\]\[y=1-4t\]\[z=2+t\] and now to CHECK if they are perpendicular, we cross product them with the point of intersection?

- Jhannybean

not cross product,i mean dot product*

- Jhannybean

we cross check them*

- Psymon

Well, you would take that cross product result and dot it with the direction vector of the original line. But it looks like you did it backwards. I think you got the negative of what I got or something O.o What I came up with works, yours doesnt. Actually, I purely got 2, 4, 1 and you had -2, -4 1.

- Jhannybean

Bah!. CROSS PRODUCTS ARE CONFUSING.

- Jhannybean

Is it... the vector found X direction vector of line?

- Jhannybean

|dw:1378713911589:dw|

- Psymon

xD
|dw:1378713833625:dw|
Not sure if you were shown how to look at doing the cross product in the correct order or not.

- Psymon

But yeah, looks like you got it right this time.

- Jhannybean

Nope, doesnt really go into detail with my book. Id forgotten that the vector from the original line had a,b,c as its components -_-

- Psymon

Trust me, Im forgettign stuff left and right, too xD My book doesnt give enough examples either. I feel like if I were quizzed on the first chapter right nowId get a C, haha. But Ive done enough of these problems to have more of an idea. But this problem and its approach is something not in my book at all, so yeah, new problem to me.

- Jhannybean

And.. you're in HS im guessing xD

- Psymon

Nah, college, lol.

- Jhannybean

k.. yeah i need to do more homework problems... D: Im using... Stewarts 7th ed Calculus.

- Psymon

Yeah, Im using some larson 10th edition one. Ive done pretty much all the odd problems in the whole chapter, so not sure what else I can do xD I just decided to review when I can and just go into the next chapter.

- Jhannybean

Eesh,.

- Psymon

I figure youll be into quadric surfaces next.

- Jhannybean

Oh we already have. Im just behind :|

- Psymon

oh x_x Yikes. Cylindrical and spherical coordinates?

- Jhannybean

Mmhmm.
Like the hyperbolic paraboloid, elliptical paraboloid, doublecones... hyperboloidof one/two sheets

- Psymon

Right. Made me feel like I was taking art class.

- Jhannybean

LOLikr....

- Jhannybean

Alrighty, i shall talk to you next time :3 Thank you for you for your help! :)

- Psymon

Alright, np and good luck xD

- Jhannybean

Thaaanks :D

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