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Stuck on vector calculus questions for exam

  1. Aug 9, 2008 #1
    two questions that i cant ever remember covering

    first of all finding a perpendicular vector

    find a vector perpendicular to the vectors a=i+2j-2k and b=-2i+3j+5k

    and secondly the equation of a plane?? through point with position vector (2,1,1) and perpendicular to (3,-1,2) what are the forumla for finding these pieces of information?
     
  2. jcsd
  3. Aug 9, 2008 #2
    Show some attempt

    1) Use cross product
    2) (3,-1,2) is perpendicular to the plane and (2,1,1) is a plane point

    These both questions are really simple (you are not extending your information etc.). If you know your basics you should be able to solve them.
     
  4. Aug 11, 2008 #3
    so the first parts just the cross product? then to find the perp unit vector i divide it by its length?
     
  5. Aug 11, 2008 #4
    whats the equation of a plane? and are there different ways to work it out given different information because the notes i have on it dont seem to make much sence to me i was expecting some kind of formula but there doesnt seem to be one
     
  6. Aug 11, 2008 #5
    when given 3 points do i do the cross product of two distances to find the normal vector? e.g A,B,C (B-A)X(C-A)=N and if this is so what do i after ive found the normal?
     
  7. Aug 11, 2008 #6

    Defennder

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    Homework Helper

    Yes correct.

    The equation of a plane is given by the dot product of a vector lying on the plane from a reference point on the plane and the vector normal to the plane.

    Yes this is correct. The rest is as explained above.
     
  8. Aug 11, 2008 #7
    anyone arround? could really use some help on this ive got to the point where i now know (or think i know) that Ax+By+Cz+D=0 is the equation of the plane with A,B, C being the normal vector and D being the distance from the plane to the orogin but i have no idea how to find these two peices of information from the information that i have been given in any of the examples, thanks
     
  9. Aug 11, 2008 #8

    Defennder

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    That is the general equation of a plane. It is obtained by the means as I have explained to you above.
     
  10. Aug 11, 2008 #9
    and secondly the equation of a plane?? through point with position vector (2,1,1) and perpendicular to (3,-1,2)

    so for this question am i correct in thinking that i need to find the line perpendicular to the perpendicular (i.e the normal) and then dot product with the other vector?
     
  11. Aug 11, 2008 #10
    but how do i find the perp of 3,-1,2 with only one vector to go on? i cant use the cross product this way as above?
     
  12. Aug 11, 2008 #11
    and for the third part i just use my normal along with any one of the 3 points and cross product them? thanks for ur help so far defender
     
  13. Aug 11, 2008 #12
    if n perp to plane

    then
    n dot any plane position = D (in standard eqn)

    and

    <a,b,c> = n

    ax+by+cz = d

    (I learned this in high school discrete math, maybe they don't teach well in calculus or assume that you know this already ..)
     
  14. Aug 12, 2008 #13
    Given a normal vector n, and a point ro, the equation of a plane perpendicular to that vector and through that point is n[tex]\circ[/tex](r- ro)=0. (That circle things represents a dot). If you are using Stewart's Calculus text, there is a pretty good section about this in that book.
     
  15. May 14, 2009 #14
    For the first one use cross product so you'll get a vector perpendicular to both
    second the d.c's of the plane are the cordinaates of the Vector perpendicular to the plane and it passes through the given point,subs and get the ans
    ax+by+cz+d=0
    a,b,c (you know them)
    it passes through
    a',b',c'
    so
    d=-(aa'+bb'+cc')
     
  16. May 14, 2009 #15
    For the first one use cross product so you'll get a vector perpendicular to both
    second the d.c's of the plane are the cordinaates of the Vector perpendicular to the plane and it passes through the given point,subs and get the ans
    ax+by+cz+d=0
    a,b,c (you know them)
    it passes through
    a',b',c'
    so
    d=-(aa'+bb'+cc')
     
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