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Find all values of x so that the triangle with vertices A=(−3, 1, x), B=(−3, 4, −2), and C=(−2, −2, 2) has a right angle at A.

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2 answers
  1. vector BA = [0, -3, x+2]
    vector CA = [-1, 3, x-2]
    if angle A = 90° then BA dot CA = 0
    0 - 9 + x^2 - 4 = 0
    x^2 = 13
    x = ±√13

    or, using Pythagoras

    AB = √(0 + 9 + (x+2) ) = √(x^2 + 4x + 13)
    AC = √(1 + 9 + (x-2)^2 ) = √(x^2 - 4x + 14)
    BC = √(1 + 36 + 16) = √53

    if angle A = 90° then
    AB^2 + AC^2 = BC^2
    x^2 + 4x + 13 + x^2 - 4x + 14 = 53
    2x^2 = 26
    x^2 = 13
    x = ±√13

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  2. thank you for the in-depth explanation

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