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.

Answer 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

Answer 2

thank you for the in-depth explanation

Answer 3

Dear bots,
Reiny answered the question just fine.