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Vector Dot Product question

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Expand and simplify:
    (2[tex]\vec{a}[/tex] + 3[tex]\vec{b}[/tex]) .(5[tex]\vec{a}[/tex] - [tex]\vec{b}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    I tried expanding it, but was a little confused, and would really appreciate any help..would it be (2[tex]\vec{a}[/tex])([tex]\vec{a}[/tex]) = 2 [tex]\vec{a}[/tex]2?

    Thanks..
     
  2. jcsd
  3. Apr 18, 2010 #2

    Mark44

    Staff: Mentor

    As your final answer? No. The dot product has a bunch of properties, some of which apply in this problem.
     
  4. Apr 18, 2010 #3
    like 10 a 2 - 3 b2

    now im confused, could you please elaborate on the properties that you mentioned before.
    thanks
     
  5. Apr 19, 2010 #4

    Mark44

    Staff: Mentor

    10a2 - 3b2 is not the right answer. What does a2 even mean?

    What's a [itex]\cdot[/itex] (b + c)?
    What's a [itex]\cdot[/itex] a?
     
  6. Apr 19, 2010 #5
    ok, so using the distributive property of Dot product, I got:

    -10a2+2 ab + 15 ab - 3b2

    Simplifying we get:

    -10a2+ 17 ab - 3b2

    is this correct?
     
  7. Apr 19, 2010 #6

    Mark44

    Staff: Mentor

    No, this is incorrect for two reasons.
    a[itex]\cdot[/itex]a [itex]\neq[/itex]a2. These are vectors and they are being multiplied using the dot product. There is a property of the dot product that allows you to do something with a[itex]\cdot[/itex]a.

    Also, two of your three coefficients are wrong.
     
  8. Apr 19, 2010 #7
    I always thought that
    a.a = lal2 (this isn't correct?)

    Two of my three coefficients are wrong? (please note that I made a mistake copying down the question in my first post, it is actually -2 a instead of 2 a). Is this correct?

    Thanks..
     
  9. Apr 19, 2010 #8

    Mark44

    Staff: Mentor

    This is correct.
    OK, then the coefficients are correct, but you shouldn't write a2, b2, or ab. All of these are dot products, and you should simplify the a.a and b.b terms.
     
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