Passing Point trouble with finding a Vector Equation of a line

  • Thread starter Liquid7800
  • Start date
  • #1
76
0

Homework Statement



The problem is stated as such:

"Find the Vector Equation of the line that forms a right angle with the lines":

L1:{(3,3,4) + t<2,2,3>}


L2:{(1,6,-1) + k<-1,2,0>}

where t and k are some scalar parameters

Homework Equations



Assuming knlowlede of vector form/notation for the equation of a line; dot product, etc.

The Attempt at a Solution



let L3 be the line intersecting L1 and L2

We begin by establishing a relation between L3 and (L1 and L2)
Since L3 is perpendicular to L1 their dot product is,

L3 *L1 = 0 OR < 2,2,3 > * < a1, a2, a3 > //// where < a1, a2, a3 > is the unknown direction vector of L3

Thus we can write the linear equation,
2a1 + 2a2 +3a3 = 0

Similarily,

L3 is perpendicular to L2 so,
L3 *L2 = 0 OR < -1,2,0 > * < a1, a2, a3 > //// where < a1, a2, a3 > is the unknown direction vector of L3

Thus,
-a1 + 2a2 = 0

With our two linear equations, we can set up a system to find the direction vector for L3 such as,

2a1 + 2a2 +3a3 = 0
-(-a1 + 2a2 = 0)
///where we take the difference here
--------------------
3a1 + 3a3 = 0

Solving for a1 we get,

a1 = -a3

Now we use a1 to get a3 in equation : -a1 + 2a2 = 0

-(-a3) + 2a2 = 0 , and solving we get

a3 = -2a2

Thus if a1 = -a3 and a3 = -2a2 then ---> the direction vector for L3 is after factoring is

---[1]---L3 : { (, , ,) + s< 2,1,-2 > }| where s = a2

---------Part B, Now to find the Passing Point,
My questions are at this point...

I ve tried setting up a system to find the intersection of say L1 and L3 to produce my 'passing point' for L3, however I cant get ANY system to produce a vaule for 's' and 't' that satisfy the equations...

For example this system produces no answer

L1 : { (3,3,4) + t<2,2,3>}
L3 : {(3 +2t, 3+2t, 4 +3t) + s <2,1,-2> }
|Where the direction vector comes from [1]

we then put
L1
x= 3 + 2t
y= 3 + 2t
z= 4 + 3t

L3
x= 3 + 2t + 2s
y= 3 + 2t + s
z= 4 + 3t - 2s


and we set them to each other---but no answer.

Ive tried various combinations to no avail such as:

3 + 2t = 1-t + 2s
3 + 2t = 6 + 2t + s
4 + 3t = -1 -2s


And I still cant get a 's' and 't' to satisfy the systems--therby producing an intersection point of L1 and L3...
The closest Ive gotten is 13 = -13 or -4 = 5 or 5 = 15...for values of s=6 and t = -3 or s = -3 and t = -8/3...etc. Ive been at this awhile so any help would be great.

I feel I am close to getting this problem...can anyone help point out what I am doing wrong--or what I am missing?

If anything is unclear let me know.

Thanks
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770

Homework Statement



The problem is stated as such:

"Find the Vector Equation of the line that forms a right angle with the lines":

L1:{(3,3,4) + t<2,2,3>}


L2:{(1,6,-1) + k<-1,2,0>}

where t and k are some scalar parameters

Homework Equations



Assuming knlowlede of vector form/notation for the equation of a line; dot product, etc.

The Attempt at a Solution



let L3 be the line intersecting L1 and L2

We begin by establishing a relation between L3 and (L1 and L2)
Since L3 is perpendicular to L1 their dot product is,

L3 *L1 = 0 OR < 2,2,3 > * < a1, a2, a3 > //// where < a1, a2, a3 > is the unknown direction vector of L3

Thus we can write the linear equation,
2a1 + 2a2 +3a3 = 0

Similarily,

L3 is perpendicular to L2 so,
L3 *L2 = 0 OR < -1,2,0 > * < a1, a2, a3 > //// where < a1, a2, a3 > is the unknown direction vector of L3

Thus,
-a1 + 2a2 = 0

With our two linear equations, we can set up a system to find the direction vector for L3 such as,

2a1 + 2a2 +3a3 = 0
-(-a1 + 2a2 = 0)
///where we take the difference here
--------------------
3a1 + 3a3 = 0

Solving for a1 we get,

a1 = -a3

Now we use a1 to get a3 in equation : -a1 + 2a2 = 0

-(-a3) + 2a2 = 0 , and solving we get

a3 = -2a2

Thus if a1 = -a3 and a3 = -2a2 then ---> the direction vector for L3 is after factoring is

---[1]---L3 : { (, , ,) + s< 2,1,-2 > }| where s = a2

Think about how much easier it would be to use the cross product of the two lines' direction vectors to get the common normal direction.

---------Part B, Now to find the Passing Point,
My questions are at this point...

I ve tried setting up a system to find the intersection of say L1 and L3 to produce my 'passing point' for L3, however I cant get ANY system to produce a vaule for 's' and 't' that satisfy the equations...

I didn't check your arithmetic for the second part, but I have a question for you. What is the definition of "passing point"? I'm guessing it isn't the same as "intersection point". Two straight lines in 3D don't have to intersect, you know.
 
  • #3
76
0
Think about how much easier it would be to use the cross product of the two lines' direction vectors to get the common normal direction.

Id like to learn how to do it this way...this is the only way my professor showed us how to do it---im curious how the cross product could get the same direction vector ---I thought normals had to do more with planes?

I didn't check your arithmetic for the second part, but I have a question for you. What is the definition of "passing point"? I'm guessing it isn't the same as "intersection point". Two straight lines in 3D don't have to intersect, you know.

My professor calls the point P0 = (x0,y0,z0) the passing point
such as P = P0 + ta | where a is a direction vector

---my professor said thats 'what he calls it'---I apologize for any confusion if there is another more commonly used term for it

I guess I dont really know how to solve the system maybe? This problem is supposed to have an intersection---at least he told us it is supposed to (dont they create these problems themselves?) and I was told if I can find an intersection of L1 and L3 or L2 and L3 I can use that intersection point as the 'passing point' to complete my equation of the line for L3
 
Last edited:
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
Id like to learn how to do it this way...this is the only way my professor showed us how to do it---im curious how the cross product could get the same direction vector ---I thought normals had to do more with planes?

If you have two vectors and think of their tails touching, they determine a plane. Their cross product, being perpendicular to each, is perpendicular to that plane.

My professor calls the point P0 = (x0,y0,z0) the passing point
such as P = P0 + ta | where a is a direction vector

---my professor said thats 'what he calls it'---I apologize for any confusion if there is another more commonly used term for it

I guess I dont really know how to solve the system maybe? This problem is supposed to have an intersection---at least he told us it is supposed to (dont they create these problems themselves?) and I was told if I can find an intersection of L1 and L3 or L2 and L3 I can use that intersection point as the 'passing point' to complete my equation of the line for L3

OK. If the two lines are given to intersect, their x, y, and z coordinates must be equal for some values of the parameters. You have the two equations given:

L1:{(3,3,4) + t<2,2,3>}

L2:{(1,6,-1) + k<-1,2,0>}

Set their x, y, and z coordinates equal. You will have 3 equations in the two unknowns t and k. Normally any two lines might not intersect and you wouldn't be able to satisfy 3 equations with only two unknowns. But if you get lucky and the lines intersect, you will find a t and k that work. Those values will give you the intersection point. Try it.
 
  • #5
76
0
Thanks for the hint...I dont know why my professor said to use L3 and (L1 OR L2) to find an intersection...I wouldve definitely tried L1 and L2...so now that my system gives me:

8/3 = 8/3 where t = -(1/6) and k = -(5/3)

I just use the intersection point from these two lines ( ) + s<2,1,-2> where my direction vector is from L3?

Is that it?
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
Thanks for the hint...I dont know why my professor said to use L3 and (L1 OR L2) to find an intersection...I wouldve definitely tried L1 and L2...so now that my system gives me:

8/3 = 8/3 where t = -(1/6) and k = -(5/3)

I just use the intersection point from these two lines ( ) + s<2,1,-2> where my direction vector is from L3?

Is that it?

That's the idea. To check your arithmetic make sure that the value of t in L1 and k in L2 give you the same point on each line. That is, make sure you get the same (x,y,z) for both lines. I got a different value for t but I may have made a mistake. Check your work.
 
  • #7
76
0
Thank you very much...I appreciate the effort in explaining things to me....x, y, and z seemed to match.

This makes sense from the perspective of a cross product graphic...but the sample graph was drawn that looked like an 'H' where L3 was the horizontal line was the line in question, so visually it didnt make sense to use L1 and L2 as having an intersection too.

Is this the ONLY way to get the P0 component (the missing component/'passing point') to complete the equation of L3
 
  • #8
76
0
Sorry to bring this thread up again, but I showed my instructor the process for
---------Part B, Now to find the Passing Point


...that LCKurtz here demonstrated...and while correct my instructor mentined that what if the lines were NOT perpendicular that intersect L3---so he discussed a more general method for find P0...such as using the given points from L1 and L2 in this manner....problem is I encounted ANOTHER error such that the system of equations are not completely satified...I should mention he only went through part of the procedure for the solution in this manner the rest he left up to me...

So,

Where point P is an element of L1 and point Q is an element of L2 we can get

if (P1, P2, P3) = (3,3,4) + t<2,2,3> AND (Q1,Q2,Q3) = (1,6,-1) + k <-1,2,4> where k = lambda in L2

So subtracting:

(Q1,Q2,Q3) = (1,6,-1) + k <-1,2,4> MINUS (P1, P2, P3) = (3,3,4) + t<2,2,3>

We get

(Q1-P1, Q2 – P2, Q3 – P3) = (-2,3,-5) +<-k-2t,2k-2t, 4k-3t> where k = lambda

So (Q1-P1, Q2 – P2, Q3 – P3) can be considered a vector for L3, which we know is <a1, a2, a3> or <2,1,-2> from the very beginning

Putting this together we have 3 equations and 2 unknowns (‘t’ and ‘k’)

So our system is:

I solved for t and k, where k is lambda in L2, and got t=-1 and k =2

2 = -2 –k -2t
1 = 3 + 2k – 2t
-2 = -5 + 4k – 3t

solving this system though does not satisfy ALL the equations (only the first) when k = 2 and t = -1---which are the values I got for k and t

I tried also substracting P from Q such that:

(P1, P2, P3) = (3,3,4) + t<2,2,3> MINUS (Q1,Q2,Q3) = (1,6,-1) + k <-1,2,4> where k = lambda in L2

And got :

<a1, a2, a3> = (2,-3,5) + <k+2t, -2k+2t, -4k+3t> where k = lambda

Getting a similar system with 2 unknowns:

2 = 2 + k+2t
1 = -3 -2k+2t
-2 = 5 -4k+3t

In this cases I got values for k = 4 and t = -2.

These values still only satisfy the first equation though…

Any clues what is going on? Thanks for any help
 
  • #9
76
0
Hmm, I rechecked some info and contrary to what I think was discussed the lines L1 and L2 do NOT intersect

....at this point something appears to be amiss, either in the problem or usually I would say the arithmetic but I have re-checked that and it seems to be okay, since for L1 and L2 by using the parametric equations set to themselves
x=x
y=y
z not= z

there may not be anything more for anyone to do but this has stumped me for some time now since it was implied L3 intersected L1 and L2
 
Top