Question

linear algebra (cross product in r^3)

find parametric equations of the line passing through (3, -1, -3) and perpendicular to the line passing through (3, -2, 4) and (0, 3, 5).

answer: x=3, y=-1+t, z=-3-5t

How do you solve this? I don't know how to get the direction vectors(?). All I know is that the direction vector of the given line is (-3, 5, 1) and you're supposed to find the perpendicular vector for that...

Answer 1

Assuming that the line you're supposed to find is also supposed to intersect with the line you were given, I have a method, but it isn't too pretty.

Answer 2

First, the direction of the first line you were given is <-3, 5, 1>. Note that any vector perpendicular to this line will lie in some plane that is perpendicular to that line. Hence, we are looking for a plane given by the equation \(-3x+5y+z+d=0\) such that this plane contains the point (3, -1, -3).

Do you follow so far?

Answer 3

Yes. But we haven't dealt with planes. We did in high school, but I'm not sure if we can solve using planes now :/

Answer 4

I'm not sure how to do this without planes. I'll finish what I have so far and see where we get though.

Answer 5

If we use the plane, we get that our equation for the plane must be \[-3(3)+5(-1)+3(-3)+d=0\]Solving for \(d\), we get that \(d=23\). So our plane equation is \[-3x+5y+z+23=0\]Now we need to find where this plane intersects the first line given to us.

Answer 6

We have that \[x=3-3t\]\[y=-2+5t\]\[z=4+t\]Substituting these into our plane equation, we get that \(t=-8/35\)

Answer 7

Now we plug this back into the equations for \(x, y, z\) to get that it intersects at the point \((129/35,\;-22/7,\; 132/35)\)

Answer 8

Using this as a second point, and \((3, -1, -3)\) as the first, we get a vector in the direction of \(<24/25, -29/7, 237/35>\)

I seem to have done something wrong :(

Answer 9

Wow those are big numbers.. I think I may have got it, but I'm not sure.
Since the direction vector is (-3, 5, 1), we can use the dot product (Assuming the perpendicular vector is (a, b, c): -3a+5b+c=0. And then can't we just plug in random numbers? So if we plug in 0 and 1 for a and b respectively, we automatically get -5 as c.

Answer 10

That would probably work better than what I tried to do. Although you probably won't necessarily get an intersection point. I guess you don't need that.

Answer 11

However, their answer for the line doesn't intersect the original line, so whatever.

Answer 12

Oh I see. Thanks for your help KingGeorge! :)

Answer 13

You're welcome.

Answer 14

I think I see where I went wrong though. When solving for \(d\), I used the equation\[-3x+5y-3z+d=0\]When I should have used\[-3x+5y+z+d=0\]Derp :(

Answer 15

Ah I see haha. So I guess it should work using the method!

Answer 16

Let's see where I get with the new numbers. Give me a minute.

Answer 17

I get a vector in the direction of \(<6/35,\;-9/7,\;243/35>\) which still isn't pretty. However, when tested with the dot product, this is perpendicular to the original line.

Answer 18

So an alternative answer (if you multiply through by 35) would be \[\vec r=\left(\begin{matrix}3\\-1\\-3\end{matrix}\right)+t\left(\begin{matrix}6\\-45\\243\end{matrix}\right)\]Which is a vector perpendicular to the given line, passes through the specified point, and also passes through the given line.

Answer 19

I guess there are a few possible answers, not just one? Thanks again :)

Answer 20

You're welcome =D