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Verify a set of vectors being a basis for a plane. How?

  1. Mar 3, 2009 #1
    Verify that B={(3,2,0),(0,2,3)} is a basis for the plane 2x[1]-3x[2]+2x[3]=0 (numbers in square brackets denote subscript)



    How to determine whether a vector, say, (1,2,2) or (5,4,1), lie in the plane given above?



    There is this hint that says to check whether the set spans the plane, solve the system that arises from asking what condition(s) an arbitrary b must satisfy to be a linear combination of the given vectors. But since I am doing not very well in this chapter, I am having some hard time getting the hint...if possible, a detailed step by step explanation of how to solve this problem will be greatly appreciated..Thanks! :)
     
  2. jcsd
  3. Mar 3, 2009 #2
    Solve the system of equations a(v1) + b(v2) = (x,y,z) in terms of a and b; you will see that one of the lines of the augmented matrix is all zeros on the left, and some expression in x, y, and z on the right; the system is consistent iff the left and right sides are equal, so you can set the rhs equal to zero.

    If this expression, set equal to zero, is the equation of the plane, then a(v1) + b(v2), with the values of a and b you found, is the point (x, y, z), and this point must lie on the plane defined by the equation above.

    E.g. v1=(1,0,0), v2=(0,1,0), and

    a(v1) + b(v2) = (x,y,z) <=> a = x, b = y, z = 0.

    z = 0 is the equation of the plane; so if I choose a = x and b = y, then a(v1) + b(v2) is equal to (x, y, 0), and is a point on the plane z = 0.
     
  4. Mar 3, 2009 #3
    Ah. So say if the vector is (1, 2, 2), I should augment it with the basis to find a and b, if possible, ie. a (3,2,0) + b (0,2,3) = (1,2,2)

    If a and b exist, then this vector lies in the plane.

    Is that correct?
     
  5. Mar 3, 2009 #4

    lanedance

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    sounds good

    this examples is also simpified by noting the plane contains the origin

    just to add a differnt spin on what was already mentioned...

    to show that B is a basis for the plane you need to show any vector in the plane can be written as a linear combination of the basis vectors, which is what you're doing with the a.v1 + b.v2

    to determine whether vectors lie in the plane quickly, can you find a vector perpindicular to the plane? then the dot product of this vector with any on the plane will be zero

    (hint: what is v1xv2 perpindicular to?)
     
  6. Mar 3, 2009 #5
    I think I am getting it alright..To truly understand the definitions and put them to use isn't quite easy after all.. :D

    What about the B-coordinates? How do I find the B-coordinates of a vector once I have verified that the vector lies in the plane?

    ps. perpendicular? I remember that it has something to do with projection, right?
     
  7. Mar 3, 2009 #6
    Wait...I have found that a is 1/3 and b is 2/3...then isn't (1/3, 2/3) the B-coordinate of the vector?
     
  8. Mar 3, 2009 #7

    lanedance

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    if v1 & v2 are your b vectors, u the vector you are trying to find, then you will have written

    v = a.v1 + b.v2

    what are the b co-ordinates, i think you mean (a,b). Gets a bit confusing if you a not using an orthogonal basis, which your basis B doesn't appear to be...

    however to find them try taking the dot product of u with v1 & then v2 then solving for a and b

    perpindicular means 90 dgree angle. The dot product give the projection of a vector onto another:
    u (dot) v = u.v.cos(theta) - (magnitudes x angle)

    which is zero if teh angle the vectors make is 90 degreees
     
  9. Mar 3, 2009 #8
    Oh, yes! I still remember how to find the perpendicular vector of a vector, and indeed it involves projection.

    As for the B-coordinates, here is how the original question stated:

    "for each of the following vectors, determine whether it lies in the plane of part a), and if so, find its B-coordinated."

    One of the "following vectors" is (1, 2, 2) and the plane is the one given in the first post..

    So the B-coordinates are the same thing as the coordinates with respect to B, right?
     
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