Help needed finding vector, parametric, symmetric equation

[agenttiny200]

Homework Statement

Find the vector, parametric and symmetric equations of a line that intersect both line 1 and line 2 at 90°.

line 1:
x = 4 + 2t
y = 8 + 3t
z = -1 - 4t

line 2:
x = 7 - 6t
y = 2+ t
z = -1 + 2t

Homework Equations

not sure. I am not asking for the answer this question, but rather a hint as to how to go about solving this. However, this doesn't mean that I would hate seeing the answer, because it will help me see if I am taking the hint the right way.

The Attempt at a Solution

 

[HallsofIvy]

I would start by finding the formula for all planes perpendicular to the first line in the form 2x+ 3y- 4z= C. Then find C such that that plane is perpendicular to the second line.

 

[agenttiny200]

now this is something that none of my teachers have explained to me: what does C do exactly? I have been told that it affects how the graph looks, but I don't understand how. and are you saying that C should be the same in both equations? or negative reciprocals?

 

[agenttiny200]

and another question: for the equation you gave above (2x+ 3y- 4z= C), shouldn't the normal vector go where you put 2, 3, and 4? or is this sort of equation different from scalar equations?

 

[LCKurtz]

Here's a couple more suggestions how you might work that problem. You can find the normal direction by crossing the two direction vectors. Call that direction $\vec n$. Then take a general point on the first line, with parameter $t$, add to it a constant times your normal vector $c\vec n$ and set that equal to a general point on your second line with parameter $s$. Looking at the components of that equation will give you three equations in the unknowns $t,\ s,\ c$. Once you know $t$ and $s$ you will know the two points your line must go through.

Alternatively you can treat it as a calculus problem. Minimize the square of the distance between general points on the first and second lines. Get $s$ and $t$ that way.

 

[agenttiny200]

so your saying this:

(t+w) x cn = (q+s)

where w is a point on line 1
where q is a point on line 2

where t, c, and s are unknowns, or the symmetric equations for x, y, or z?

And I am guessing that the t is the unknown from the parametric equations I gave for line 1,
s is the unknown from the parametric equation for line 2 if they are unknowns.

 

[LCKurtz]

so your saying this:

(t+w) x cn = (q+s)

where w is a point on line 1
where q is a point on line 2

where t, c, and s are unknowns, or the symmetric equations for x, y, or z?

And I am guessing that the t is the unknown from the parametric equations I gave for line 1,
s is the unknown from the parametric equation for line 2 if they are unknowns.

That is very confused. Your two lines are$$
\vec R_1 = \langle 4,8,-1\rangle + t \langle 2,3,-1\rangle$$ $$
\vec R_2 = \langle 7,2,-1\rangle + s\langle -6,1,2 \rangle$$Now calculate the normal vector and try what I suggested.

 

[agenttiny200]

I understood that part, but what I don't get is what your doing with the normal vector. from your earlier explanation, it seems as though [4,8,-1] is being multiplied by the normal.

 

[LCKurtz]

I understood that part, but what I don't get is what your doing with the normal vector. from your earlier explanation, it seems as though [4,8,-1] is being multiplied by the normal.

"Your" is a possessive pronoun. If you mean "you are" you use the contraction "you're".

Have you calculated the normal yet? I'm waiting for you to actually try my suggestion.

 

[agenttiny200]

dude, I have spent the last three weeks doing quite a bit of finding normals:
[2,3,-4] x [-6,1,2] = [(3)(2)-(1)(-4),(-4)(-6)-(2)(2),(2)(1)-(-6)(3)]
= [6+3,20+4,2+18]
= [9,24,20]

And there is the normal.

but as I said, the one thing I don't understand from the explanation you gave is how you use the normal vector (I am very weak in word problems, and I have tried to change that with little avail)

 

[LCKurtz]

dude, I have spent the last three weeks doing quite a bit of finding normals:
[2,3,-4] x [-6,1,2] = [(3)(2)-(1)(-4),(-4)(-6)-(2)(2),(2)(1)-(-6)(3)]
= [6+3,20+4,2+18]
= [9,24,20]

And there is the normal.

but as I said, the one thing I don't understand from the explanation you gave is how you use the normal vector (I am very weak in word problems, and I have tried to change that with little avail)

Apparently you are a little weak on your arithmetic too, since you have 2 of the 3 components wrong. You should get [10,20,20]. And to keep things simple you can use $\vec n =\langle 1,2,2\rangle$. Now you want $\vec R_1 +c\vec n = \vec R_2$. This gives$$
\langle 4,8,-1\rangle + t \langle 2,3,-4\rangle + c\langle 1,2,2\rangle = \langle 7,2,-1\rangle + s\langle -6,1,2 \rangle$$This says some point on the first line plus some multiple of the normal has to get you some point on the second line. You have three unknowns $t, c, s$ and three equations by equating components.

 

[agenttiny200]

okay, ill try that out and get back to you.

 

[NATURE.M]

Does it matter whether you use (1) R1-R2=cn or (2) R1+cn=R2 (the form which you've outlined).
I ask this because I've seen other individuals use (1), and final answers vary. As (1) provides you with s=1, t=-1 and n=1, and I used (2) and which provides me with s=-1...so which method is right? or are they both right?
(Please don't mind my notation).

 

[LCKurtz]

Does it matter whether you use (1) R1-R2=cn or (2) R1+cn=R2 (the form which you've outlined).
I ask this because I've seen other individuals use (1), and final answers vary. As (1) provides you with s=1, t=-1 and n=1, and I used (2) and which provides me with s=-1...so which method is right? or are they both right?
(Please don't mind my notation).

This thread is almost a year old and the OP disappeared long ago. Also it is not clear to whom you are replying or to what you are following up. If you have a topic you wish to discuss please start a new thread.