1. Questions

vectors

If |x| = 11, |y| = 23, and |x-y| = 30, find |x+y|.

Textbook answer: 20

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Iris

  1. x = a i + b j
    y = c i + d j

    a^2 + b^2 = 11^2 = 121
    c^2 + d^2 = 23^2 = 529

    x-y = (a-c)i + (b-d)j
    (a-c)^2 + (b-d)^2 = 30^2 = 900 = a^2 - 2ac + c^2 + b^2 -2bd + d^2
    so 900 = 121+529 -2(ac+bd)
    so 250 = -2(ac+bd)
    now
    x+y = (a+c)i + (b+d)j
    |x+y|^2 = a^2+2ac +c^2 +b^2+2bd+d^2 = 121+529 +2(ac+bd)
    but we know 2(ac+bd) = -250
    so
    |x+y|^2 = 121+529-250 = 400
    so in the end
    |x+y| = 20
    Whew !!!

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    Damon

  2. Thank you so much :)))

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    Iris

  4. or ...
    make a sketch of a triangle with sides 11, 23 and 30
    side 11 would be vector x, 23 would be vector -y and 30 is the resultant x-y
    by the cosine law: 30^2 = 11^2 + 23^2 - 2(11)(23)cosØ
    cosØ = -250/506

    reflect vector -y and draw in vector y. Complete the new triangle with a resultant of x+y
    |x+y|^2 = 11^2 + 23^2 - 2(11)(23)cos (180-Ø) , but cos(180-x) = -cosx = 250/506
    = 650 - 2(11)(23)(250/506)
    = 650 - 250
    = 400
    |x+y| = 20

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    Reiny

  5. Somehow you can tell one of us does physics and the other does math :)

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    Damon

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