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Zara26

find the components of the vector 3i+2j+8k in the direction of the vector 2i+2j+2k

  • one year ago
  • one year ago

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  1. sam30317
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    3i+2j+8k +2i+2j+2k ______________ 5i +4j +10k Check if its right. Been a looong time since ive done this.

    • one year ago
  2. dumbcow
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    im guessing find magnitude of 1st vector --> sqrt(3^2 + 2^2+8^2) = sqrt(77) and then find components of directional vector with same magnitude unit vector of 2nd vector = 1/sqrt3 , 1/sqrt3 , 1/sqrt3 multiply by magnitude --> sqrt(77/3) i + sqrt(77/3) j + sqrt(77/3) k

    • one year ago
  3. Zara26
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    r u sure about the answer

    • one year ago
  4. sam30317
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    .....NO!!! Im not confident thats right. I suggest asking this question in the physics study group. And look at t his site: http://www.ehow.com/how_8396057_determine-resultant-vector.html

    • one year ago
  5. dumbcow
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    im pretty sure i did it right as long as i understood the question correctly. you taking 1st vector and placing it so its going same direction as 2nd vector but has same magnitude...correct?

    • one year ago
  6. Zara26
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    am not sure...am still trying to solve it

    • one year ago
  7. sam30317
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    R u doing college physics?? Did u read anything about "dot product"??

    • one year ago
  8. wasiqss
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    dumb cow is correct ,

    • one year ago
  9. wasiqss
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    wait, ik the right answer.

    • one year ago
  10. wasiqss
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    we take the dot product of first vector with unit vector of second

    • one year ago
  11. sam30317
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    Instead of adding, do what I showed you in the SAME set up, except u multiply straight down: (3*2)+(2*2)+(8*2) = 26

    • one year ago
  12. sam30317
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    Yeaaah!! I dont remember how to get angle..if its needed

    • one year ago
  13. wasiqss
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    zara my answer is the right one, cause i did this question last day

    • one year ago
  14. Zara26
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    thankss

    • one year ago
  15. FoolForMath
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    \(<3,2,8> . <2,2,2> = 6 + 4+16 = 26\) \[|<2,2,2>|= \sqrt{12} \] Vector component of \(\vec{a} \)in the direction of \(\vec{b}\) is given by \( \frac {\mathbf{a} \cdot \mathbf{b}} {|\mathbf{b}| } \frac {\mathbf{b}} {|\mathbf{b}|} \) Hence, \(\frac {26}{12} <2,2,2> = \frac {13}6<2,2,2> \)

    • one year ago
  16. FoolForMath
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    I have assumed that you meant vector component and not the scalar one.

    • one year ago
  17. FoolForMath
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    @dumbcow: I am not sure from where you came up with that definition, can you please explain? http://en.wikipedia.org/wiki/Vector_projection

    • one year ago
  18. anonymoustwo44
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    foolformath's right

    • one year ago
  19. dumbcow
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    @foolFormath, oops did not recognize this as vector projection...for some reason i thought the magnitude of 1st vector should remain the same

    • one year ago
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