The turnbuckle is tightened until the tension in the cable AB equals 2.9 kN. Determine the vector expression for the tension T as a force acting on member AD. Also find the magnitude of the projection of T along the line AC.
The Attempt at a Solution
I was able to find the vector expression for T without any real trouble:
AB = <2.6, 1.7, -6.1>
|AB| = sqrt(2.62+ 1.72 + (-6.1)2)
T*AB/|AB| = 0.42364<2.6, 1.7, -6.1>
T = 1.101i + 0.7201j - 2.584k
I know this part is correct but I'm having difficulty doing the projection.
I tried taking taking the dot product of the T vector with the unit vector AC:
AC = ,2.6, -2.8, -6.1>
|AC| = sqrt(2.62 + (-2.8)2 + (-6.1)2)
|AC| = 7.1979
AC/|AC| = 0.3612, -0.389, 0.8475
T°AC/|AC| = 1.101*0.3612 +(-0.389*0.7201)+(-0.8475*-2.584)
But it says my answer is wrong and I'm not sure where I'm going wrong.
Advice would be greatly appreciated.