1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Dot product

  1. Sep 13, 2006 #1
    Nice, simple, straighforward problem:
    [itex]j \cdot {-j}[/itex]

    Difficulty: I solved it two ways and got two answers.

    [itex]\vec{a} \cdot \vec{b}=ab\cos\phi[/itex]

    So in this case, I should get [tex]j \cdot {-j}=(1)(-1)(cos(180^o))=(1)(-1)(-1)=1[/itex]

    However, using the formula [itex]\vec{a} \cdot \vec{b}=a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}[/itex], I got [itex]j \cdot {-j}=a_{y}b_{y}=(1)(-1)=-1[/itex]

    Can anyone tell me why?
    Last edited: Sep 13, 2006
  2. jcsd
  3. Sep 13, 2006 #2
    Recall that the dot product is:
    [tex] \vec A \cdot \vec B = |\vec A| |\vec B| \cos \theta_{AB} [/tex]

    let [itex] \vec A = (0,-1,0) [/itex]
    [tex] A=|\vec A| = \sqrt{(0)^2 +(-1)^2 +(0)^2} [/tex]

    What is [itex] (-1)^2 [/itex] :)
  4. Sep 13, 2006 #3
    Neither my book nor my professor mentions the absolute value thing. I guess that's completely implied, though, because the negative sign has to do with direction, not magnitude.

    Thank you so much!
  5. Sep 13, 2006 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    [tex] |\vec A|[/tex] means magnitude of [tex] \vec A[/tex], which is a scalar and is given by the square root of the sum of the squares of the magnitudes of the component vectors. [tex]|\vec{a}|\,=\,\sqrt{a^2_x+a^2_y+a^2_z}[/tex]

    One could also use the fact that -j * j = - (j * j) = - (1) = -1, where I use * to mean the dot product.

    Presumable j is a unit vector (0, 1, 0).

    Back to the OP:

    [tex]j \cdot {-j}=(1)(1)(cos(180^o))=(1)(1)(-1)=-1[/tex]
    Last edited: Sep 13, 2006
  6. Sep 13, 2006 #5
    I think basically your mistake lies in your calculation of the modulus(magnitude) of -j which is 1 not -1.For use in the formula a*b=abcos#
    where * means dot product
    a/b mean the magnitudes(moduli) of a & b
  7. Sep 13, 2006 #6
  8. Sep 13, 2006 #7
    The dot product operation has an associative property, that says:

    [tex] (k \vec A) \cdot \vec B = k (\vec A \cdot \vec B) [/tex]
    Where [itex] k [/itex] is a scalar, and [itex] \vec A, \,\, \vec B [/itex] are vectors.
    Last edited: Sep 13, 2006
  9. Sep 13, 2006 #8
    I've got to start consistently thinking of a negative sign as -1 multiplied by what follows, don't I.
  10. Sep 14, 2006 #9
    I remember how much grief that used to give me :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook