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Prooving 2 Vector Eqn Lines Are Perpendicular

  1. Sep 18, 2006 #1
    Hey, just hoping sum1 new a better way to do the follwing Q other then the way i'm doin it!!!

    it gives 2 vector eqn's of lines L1 and L2 as follows:

    L1 = r = (2i+4j)+k(-1i+3j)
    L2= r = (-3i+1j)+m(6i+2j)

    (ie k= lambda and m=mew- also they are inform matrice form but i didnlt now how to type i into that format so sorry for that)

    i was converting them to cartesian eqn y=mx+c, then the gradients must multiply =-1......howveer i suspect u can use the scalar product rule (ie a.b = 0 for perpendicular lines) but i can seem to work the equations to give me = 0....any assistance or hints would be greatly appreciated!!!
  2. jcsd
  3. Sep 18, 2006 #2
    i cant* cant* lol eem to work it to giv me = 0
  4. Sep 18, 2006 #3


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    Go through the scalar product rule again, maybe you accidentaly mixed up some vector components or so (not that I'm underestimating you, don't get me wrong :wink: ).
  5. Sep 18, 2006 #4


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    With the lines, think about what part of the equations affect the angle between them. Then find the scalar product of that certain part of the equations.
  6. Sep 18, 2006 #5
    i still keep getting a wole lot of mews and lambdas which dont cancel to give 0???? what do you mean by the part that effects the angle between them....given the intially position vector and then adding the variable vector...shoulnt the angle be the same regardless of value of mew and lambda as this just should extend the line...??? sorry i dont get wat you mean
  7. Sep 18, 2006 #6


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    ok. If you have two lines, such as shown below:

    The two lines make an angle with eachother. The equations of each line are shown in the top right hand corner.

    If you keep the gradient of each line the same, but move it up/down and left/right, like so:


    Notice how the angle between the lines is exacly the same. You could then say that the only thing that affects the angles between two lines is the gradient of each.

    In those vector equations, think about what part is the gradient. And then see if you can work from there.
  8. Sep 18, 2006 #7
    dam ur good at explainin!!! lol
  9. Sep 18, 2006 #8


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    So you understand the question now? :smile:
  10. Sep 18, 2006 #9
    dam...how com is so easy when u put it like that!!!

    simply usin a.b= 0 for perpendicular lines and given gradient will be determining factor for angle

    thanks heaps...really helped
  11. Sep 18, 2006 #10


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    Glad it makes sense now :smile: Glad to be of assistance :smile:
  12. Sep 18, 2006 #11
    whilst ur still there with the following...almost the same Q.....fincd the vector equation of the line perpendicular to:
    vecotr eqn r = (2i-j)+(lambda)(3i+2j) and passing through point a,position vector (4i+3j).

    My Q is, i get the gradient down to either (2i-3j) or (-2i+3j), with the answer being the latter, how do you make th decision that the latter is the corret combination...???
  13. Sep 18, 2006 #12


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    Shouldnt matter i would think.

    Look at it like this.


    See the red point at <4,3>? Try going left 2 units, and up 3, and draw a line between the two points <-2,3>. Now try going right 2 units, and down 3, <2,-3>, and again draw a line. The lines will align with eachother, giving the same result :)
  14. Sep 18, 2006 #13
    thanks again....thats what i thought (ie that they wil both be perpendicular...ie they seem to be the same line) just checkin since answer only gave one...but thanks again!!:)
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