Math Is Fun Forum

A.R.

Guest

Linear Algebra

The question is:

Find a parametric equation for the line L which passes through A and is perpendicular to both L1 and
L2.

L1: x = 1, y = 4 + 3t, z = 2 − t
L2: x = 2 + t, y = −t, z = 1 − 2t

and point A = (1, 1, 3)

It would be nice if someone could point me in the right direction. Thanks!

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Linear Algebra

The line that is perpendicular to L1 and L2 is found by taking the vector product of those two.
Once you've got that, I believe you can change the constant in each bit of the resulting vector into 1, 1 and 3. Hopefully that should shift the line where you want it, without changing its direction.

---

Why did the vector cross the road?
It wanted to be normal.

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Linear Algebra

to find a vector perpendicular to two others in R³ we use the cross product.

so one solution is:

similarly since t can be any value you can equally have

etc.

Last edited by luca-deltodesco (2008-01-21 08:24:57)

---

The Beginning Of All Things To End.
The End Of All Things To Come.

A.R.

Guest

Re: Linear Algebra

Thanks for the replies.

Is it possible to get the answer without using the cross product? We havn't covered that in class yet so I would assume it wouldn't be involved in the answer.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Linear Algebra

Well, two vectors are perpendicular to each other if their dot product is 0.
Therefore, if your new line is L then L . L1 = 0 and L . L2 = 0.

Saying that L is (x,y,z), you can find two equations that are equal to 0, and then solve them simultaneously to find L.

It's more long-winded, but if you're not allowed cross products then I think this is the best way.

---

Why did the vector cross the road?
It wanted to be normal.

A.R.

Guest

Re: Linear Algebra

Well, two vectors are perpendicular to each other if their dot product is 0.
Therefore, if your new line is L then L . L1 = 0 and L . L2 = 0.

Saying that L is (x,y,z), you can find two equations that are equal to 0, and then solve them simultaneously to find L.

It's more long-winded, but if you're not allowed cross products then I think this is the best way.

So then,

I only have to worry about the vectors v, and not about the two other vectors/points (in this case, [1,4,2] and [2,0,1]), right?

So it would look like,

[x,y,z] (dot) [0,3,-1] = 0
[x,y,z] (dot) [1,-1,-2] = 0

3y - z = 0
x - y - 2z = 0

And then solve as a set of lin. eqns?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Linear Algebra

Precisely! You don't need to worry about where the vectors are, only what direction they're pointing in. After you've found the perpendicular vector, you're going to be shifting its position to fit with your other condition anyway.

It's also worth noting that since you have three variables but only two equations, you're going to end up with infinite solutions. This is good, because it's consistent with what happened when luca used cross products. You just need to set x (for example) equal to 1 and then put a parameter outside the vector afterwards.

---

Why did the vector cross the road?
It wanted to be normal.

A.R.

Guest

Re: Linear Algebra

Precisely! You don't need to worry about where the vectors are, only what direction they're pointing in. After you've found the perpendicular vector, you're going to be shifting its position to fit with your other condition anyway.

It's also worth noting that since you have three variables but only two equations, you're going to end up with infinite solutions. This is good, because it's consistent with what happened when luca used cross products. You just need to set x (for example) equal to 1 and then put a parameter outside the vector afterwards.

Allright, thanks so much for the help.  Makes a LOT more sense now.