Physics

posted by Ismail

A and B are vectors in the same plane. If A= 8i+6j. Find the vector B such that |A+B| = 14.14 units and A is perpendicular to B.

  1. Damon

    A is perpendicular to B
    so
    A dot B = 0
    AxBx + AyBy = 0
    8 Bx + 6 By = 0
    and
    A + B = (Ax + Bx)i+ (Ay+By)j
    |A+B|^2 = 14.14^2 = (Ax+Bx)^2+(Ay+By)^2
    so
    (Ax+Bx)^2+(Ay+By)^2 = 100
    (8+Bx)^2 +(6+By)^2 = 100
    but we know
    By =-(8/6)Bx = -(4/3) Bx
    plug and chug.

  2. Ismail

    Sorru for the inconvenience... but I didnt get this concept
    |A+B|^2 = 14 ??!!

  3. Damon

    |A+B|^2 = 14.14^2 = 200 not 100

    I did not use calculator, knew sqrt 2 = 1.414

  4. Ismail

    Oh .. the rest of the steps are the same then ?

  5. Damon

    yes

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