Coordinates of the points A, B and C are (1, 2, 3) , (2, 3, -1) and (3, -1, 1) respectively,
find vectors ->AB and ->AC
find the angle <BAC
to the nearest degree:
please help with this problem
Well, vector AB will be:
(2 - 1)i + (3 - 2)j + (-1 - 3)k = i + j - 4k
Vector AC will be:
(3 - 1)i + (-1 - 2)j + (1 - 3)k = 2i - 3j - 2k
(where i, j, and k are the unit vectors in the x, y, and z directions, respectively.)
We also know that
(AB) (dot) (AC) = |AB|*|AC|* cos(theta)
where theta is the angle between the vectors.
(AB) (dot) (AC) = (1)*(2) + (1)*(-3) + (-4)*(-2) = 7
|AB| = sqrt{1^2 + 1^2 + 4^2} = sqrt{18} = 3*sqrt{2}
|AC| = sqrt{2^2 + 3^2 + 2^2} = sqrt{17}
So cos(theta) = 7/(3*sqrt{2}*sqrt{17})
Now take the inverse cosine to get the angle.
-Dan