- #1

Liquid7800

- 76

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## Homework Statement

The problem is stated as such:

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

L

_{1}:{(3,3,4) +

*t*<2,2,3>}

L

_{2}:{(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 L

_{3}be the line intersecting L

_{1}and L

_{2}

We begin by establishing a relation between L

_{3}and (L

_{1}and L

_{2})

Since L

_{3}is perpendicular to L

_{1}their dot product is,

L

_{3}*L

_{1}= 0 OR < 2,2,3 > * < a

_{1}, a

_{2}, a

_{3}>

*//// where < a*

_{1}, a_{2}, a_{3}> is the unknown direction vector of L_{3}Thus we can write the linear equation,

2a

_{1}+ 2a

_{2}+3a

_{3}= 0

Similarily,

L

_{3}is perpendicular to L

_{2}so,

L

_{3}*L

_{2}= 0 OR < -1,2,0 > * < a

_{1}, a

_{2}, a

_{3}>

*//// where < a*

_{1}, a_{2}, a_{3}> is the unknown direction vector of L_{3}Thus,

-a

_{1}+ 2a

_{2}= 0

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

_{3}such as,

2a

_{1}+ 2a

_{2}+3a

_{3}= 0

-(-a

_{1}+ 2a

_{2}= 0)

*///where we take the difference here*

--------------------

3a

_{1}+ 3a

_{3}= 0

Solving for a

_{1}we get,

a

_{1}= -a

_{3}

Now we use a

_{1}to get a

_{3}in equation : -a

_{1}+ 2a

_{2}= 0

-(-a

_{3}) + 2a

_{2}= 0 , and solving we get

a

_{3}= -2a

_{2}

Thus if a

_{1}= -a

_{3}and a

_{3}= -2a

_{2}then ---> the direction vector for L

_{3}is after factoring is

---[1]---L

_{3}: { (, , ,) + s< 2,1,-2 > }| where s = a

_{2}

*---------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 L

_{1}and L

_{3}to produce my 'passing point' for L

_{3}, however I can't get ANY system to produce a vaule for 's' and 't' that satisfy the equations...

For example this system produces no answer

L

_{1}: { (3,3,4) + t<2,2,3>}

L

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

we then put

L

_{1}

x= 3 + 2t

y= 3 + 2t

z= 4 + 3t

L

_{3}

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 can't get a 's' and 't' to satisfy the systems--therby producing an intersection point of L

_{1}and L

_{3}...

The closest I've 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. I've 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