find, correct to the nearest degree the three angles of a triange with vertices A(0,3,3) B(-2,1,1) C(1,5,-1)
The line segment AB is parallel to the vector < -2-0,1-3,1-3 > = < -2,-2,-2 >
The line segment AC is parallel to the vector < 1-0,5-3,-1-3 > = < 1, 2,-4 >
The line segment BC is parallel to the vector < 1-(-2),5-1,-1-1 > = < 3, 4,-2 >
We use this formula
Angle A is the angle between AB and AC, which is the same angle
as the angle between the vectors < -2,-2,-2 > and < 1, 2,-4 >
Their dot product is
< -2,-2,-2 >@< 1, 2,-4 > = (-2)(1)+(-2)(2)+(-2)(-4) = -2-4+8 = 2
The norm of < -2,-2,-2 > is
The norm of < 1,2,-4 > is
Substituting in the formula for the cosine,
=.1259881577
So A = 82.76217579° or 83° to the nearest degree.
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Find the other two angles the same way.
Edwin