Vectors

A triangle has vertices A(2,3,7), B(0,-3,4) and C(5,2,-4)
A) determine the largest angle in the traingle
b) determine the area of of the triangle

pls tell me how to solve the problem

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asked by Marie
  1. The largest angle is opposite the largest side.

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    posted by Damon
  2. a opposite angle A etc
    a^2 = 5^2 + 5^2 + 8^2 = 114
    a = 10.7
    b^2 = 3^2 + 1^2 +11^2 = 131
    b = 11.4
    c^2 = 2^2 + 6^2 + 3^2 = 49
    c = 7
    so side b is biggest

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    posted by Damon
  3. let AB represent vector AB, and AC as vector AC
    AB = (-2,-6,-3), |AB| = 7
    AC = )3,-1,-11) |AC| = √131
    AB•AC = |AB||AC|cos Ø
    cosØ = 33/(7√131)
    Ø = 65.676

    area = (1/2)(AB)(AC)sinØ
    = 36.5

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    posted by Reiny
  4. Now let's see if one of those angles is 90 degrees because that would make the area easy.
    dot product is magnitudes *cos angle
    AB = -2 i -6 j -3 k
    AC = +3 i -1 j -11k
    AB dot AC = -6+6 +33 = +33
    we already know |AB|=c = 7, |AC|=b = 11.4
    so
    7*11.4* cos A = 33
    cos A = .4135
    A = 65.6 degrees
    Now find angle B
    BA = -AB so
    BA = 2 i + 6 j + 3 k
    BC = 5 i + 5 j - 8 k
    BA dot BC = 10 + 30 - 24 = 16
    7*10.7* cos B = 16
    cos B = .2136
    B = 77.7 degrees
    Oh well not a right triangle
    We could find angle C by subtraction from 180 but calculate it as a check.
    CA = - AC
    CA = -3 i + 1 j +11 k
    CB = -5 i - 5 j + 8 k
    10.7*11.4 cos C = 15-5+88 =98
    cos C = .8034
    C = 36.5 degrees
    check
    36.5 + 77.7 + 65.6 = 179.8 which is close enough to 180
    Now you have three sides and the opposite angles. I will leave it to you to find the area.

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    posted by Damon

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