1. Not finding help here? Sign up for a free 30min 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!

Physics 221 Calc-based question for homework

  1. Jan 20, 2017 #1
    1. The problem statement, all variables and given/known data
    If vector d1--> is equal to 2i^-4j^+9k^ and vector d2-> is equal to 8i^-10j^+9k^, what is (d1-->+d2-->) * (d1--> X 4d2->)?

    2. Relevant equations


    3. The attempt at a solution
    So I added up first two vectors and then multiplied it by the cross product but I get a vector and when I find its length its like 1931. I just want to know if this is even close.
     
  2. jcsd
  3. Jan 20, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi AidenPhysica, Welcome to Physics Forums.

    Your result does not look correct to me. Can you show the details of your calculations?
     
  4. Jan 20, 2017 #3

    DrClaude

    User Avatar

    Staff: Mentor

    Your equation is her to read. Is the problem to calculate
    $$
    (\vec{d_1} + \vec{d_2}) \cdot (\vec{d_1} \times 4 \vec{d_2})
    $$
    ?

    If that is correct, then your result is wrong. What to you get when you take the cross product of two vectors?
     
  5. Jan 20, 2017 #4
    Its right what you put except there has to be a 4 right in front of d2 so its multiplied by four.
     
  6. Jan 20, 2017 #5

    DrClaude

    User Avatar

    Staff: Mentor

    I've corrected it. So, what do you get when you take the cross product of two vectors?
     
  7. Jan 20, 2017 #6
    Ok, so this is what I did. I first added up the two vectors to get 10i+-14j+18k. Then you do the cross product, so I did that operation of the cross product, which is basically how much two vectors are perpendicular to each other? So here is the matrix i j k on top second row is 2, -4, 9. Third row is 32i-40j+36k. The cross product of this matrix is
     
  8. Jan 20, 2017 #7
    216i +216j+48k
     
  9. Jan 20, 2017 #8

    DrClaude

    User Avatar

    Staff: Mentor

    I'll give you a hint: the actual values of the vectors are not important.

    What is the orientation of that vector with respect to the original vectors?
     
  10. Jan 20, 2017 #9
    Then what I did was I multiplied each respective component of the first answer times the second to get 3815 after doing pythagrean theorem is that better?
     
  11. Jan 20, 2017 #10
    OK, so what do you mean by orientation?
     
  12. Jan 20, 2017 #11

    DrClaude

    User Avatar

    Staff: Mentor

    Stop a minute. I want you to think about the problem first. You can calculate numerical values later.

    The direction in which it points. When you take the cross product of two vectors, the resulting vector points in a very specific direction with respect to the two other vectors.
     
  13. Jan 20, 2017 #12
    so isn't the cross product equal to magnitude of a times magnitude of b multiplied by sin of theta?
     
  14. Jan 20, 2017 #13
    Isn't cross product equal to magnitude of a times magnitude of b times sin of theta?
     
  15. Jan 20, 2017 #14
    so isn't it just: ok so we have 10.05 times 26.83 times sin of theta=309.22. So then theta=309.22/269.6*(arcsin)
     
  16. Jan 20, 2017 #15
    I don't know it doesn't work.
     
  17. Jan 20, 2017 #16
    I try plugging it into a times b sin theta = magnitude of cross product but it just doesn't work.
     
  18. Jan 20, 2017 #17

    DrClaude

    User Avatar

    Staff: Mentor

  19. Jan 20, 2017 #18
    so how do i find orientation though?
     
  20. Jan 20, 2017 #19
    Is it in degrees? I am just confused I see the picture, its like perpendicular.
     
  21. Jan 20, 2017 #20

    DrClaude

    User Avatar

    Staff: Mentor

    Yes, the resulting vector is perpendicular to the plane of the two initial vectors.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Physics 221 Calc-based question for homework
Loading...