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Cross Product and Dot Product

  1. Apr 26, 2007 #1
    Use two methods to determine a unit vector perpendicular to both (2,1,-3) and (-1, 7, 4)

    Using Cross Product:
    Let v be a perpendicular vector to the given vectors

    v = a*b
    = (2,1,-3)(-1,7,4)
    =((1)(4)-(7)(-3), (-3)(-1)-(4)(2), (2)(7)-(-1)(1))
    =(4+21, 3-8, 14+1)
    =(25, -5, 15)
    = (5, -1, 3)
    = |(5,-1, 3)|
    = sqr(25+1+9

    Let u be a unit vector in the direction of (5, -1, 3)
    u = v/|v| = 1/sqr(35)(5, -1, 3) = (5/sqr(35), -1/sqr(35), 3/sqr(35))

    Using dot product:
    Let p = (x, y, z) represent the vector that is perpendicular to the vectors.
    a = (2, 1, -3) and b = (-1, 7, 4)
    Then p*a = 0 and p*b = 0
    (x, y, z)(2, 1, -3) = 0 and (x, y, z)(-1, 7, 4) = 0

    2x + y - 3z = 0
    -x + 7y + 4Z = 0

    Isolate x in equation (1) to get x = -y/2 + 3z/2
    Substitute this into equation (2)

    -(-y/2 + 3z/2) + 7y +4z = 0
    y/2 - 3z/2 +7y +4z = 0

    z = 0 --> substitute into equation (1) gives 2x + y = 0 and so x = -y/2

    If you let x = 1, then y = -2

    Therefore p = (x, y, z) = (1, -2, 0), which gives a different vector from the one found using cross product.

    Determine the magnitude of this vector and then find a unit vector as required.

    |(1, -2, 0)| = sqr(1+4) = sqr(5)

    Let u be a unit vector in the direction of (1, -2, 0)

    u = v/|v| = 1/sqr(5)(1, -2, 0) = (1/sqr(5), -2/sqr(5), 0)

    The two answers should match up but they don't and I am not sure what I am doing wrong. I have been over it a number of times but I just can not see where the error lies. Any help would be greatly appreciated
  2. jcsd
  3. Apr 26, 2007 #2

    D H

    Staff: Mentor

    What justifies this step?
  4. Apr 26, 2007 #3
    You can substitute in any value for z, however 0 would be the only value that would isolate x and y. At least that is how I understand it from the sample equations.
  5. Apr 26, 2007 #4

    D H

    Staff: Mentor

    You had it correct to this point. These are two equations in three unknowns. One solution is the trivial solution, x=y=z=0. There are an infinite number of other solutions because two equations in three unknowns do not yield a unique solution.

    You cannot arbitrarily set one of the variables (say z) to zero because then you will find that the other two variables (x and y) must also be zero. This is the trivial solution, and it is the one solution you do not want.

    What you can do is to express any two of the variables in terms of the third. For example, find a way to express x and y in terms of z. There is an implied third equation: The desired vector is a unit vector. You can use this to come up with a value for z and hence for x and y. Note: the sign of the value for z will be arbitrary, as the unit vector equation will result in something like [itex]z^2 = a[/itex]. You will get identical results if you choose x or y as the basis variable instead of z.
  6. Apr 26, 2007 #5
    I'm sorry but I have been staring at this question for close to three hours now and I still no closer to resolving it as when I started.

    How exactly is the fact that the vector is a unit vector going to help me determine a value for z?
  7. Apr 26, 2007 #6


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    Staff Emeritus
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    Gold Member

    Basically, it means try to solve it again, only this time without z=0. You'll get a vector that looks something like

    (z+2, 3z, z) (I didn't actually work it out, but it should have one independent parameter).

    You then have to normalize this by finding a value of z such that the sum of the square of the terms equals one
  8. Apr 26, 2007 #7

    D H

    Staff: Mentor

    You started with a general form for the normal vector [itex]\vec n = (x,y,z)[/itex] and then developed two equations for [itex]x,y,z[/itex]. From those two equations, you should be able to reduce those equations to something of the form [itex]x=a*z, y=b*z[/itex] where [itex]a[/itex] and [itex]b[/itex] are constants. With this, the normal vector becomes [itex]\vec n = (a*z, b*z, z)[/itex].

    You want this to be a unit vector. What is the inner product of a unit vector with itself? What is the inner product of [itex]\vec n[/itex] with itself?
  9. Apr 26, 2007 #8

    D H

    Staff: Mentor

    This is not a three-hour problem. You had it right up to this point:

    Here is where you went wrong:

    What you should have done was to work with this equation:
    y/2 - 3z/2 +7y +4z = 0
    combining terms,
    15/2 y + 5/2 z = 0
    y = -z/3
    With this it should be obvious that choosing z=0 forces y (and hence x) to be zero as well. So choosing z=0 was exactly the wrong thing to do.

    Can you finish with y=-z/3 (alternatively, z=-3y) and your equation for x in terms of y and z?
  10. Apr 26, 2007 #9
    Ok, I think I have the idea. z = -3y
    x = -y/2 + 3(-3y)/2 = -10y/2 = -5y
    Therefore, (-5y, y, -3y)
  11. Apr 26, 2007 #10

    D H

    Staff: Mentor

    Excellent. Now finish it. You want a unit vector.
  12. Apr 26, 2007 #11


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    Homework Helper
    Gold Member

    Yes. And after normalizing to a unit vector, the "y" drops out completely.

    Notice that what you get is exactly minus th evector you got from the other method so it's not the same vector! But that's ok because there are actually *two* unit vectors perpendicular to [itex] \vec{a} [/itex] and [itex] \vec{b} [/itex]. You found one by the first technique and you found the other one by the second technique
  13. Apr 26, 2007 #12
    Thanks for all the hand holding guys. I am doing this course by correspondence and sometimes the text is less than ideal.
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