Answer
$\theta = 31°$
Work Step by Step
To calculate the angle we must first determine the following:
$r_{AB} = [-3i -0.75j + k]\ m$
$r_{AB}' = \sqrt{(-3)^2 + (0.75)^2 + (1)^2} = 3.25\ m$
$r_{AC} = [-3i + j + 1.5k]\ m$
$r_{AC}' = \sqrt{(-3)^2 + (1)^2 + (1.5)^2} = 3.50\ m$
The dot product of the two vectors is:
$r_{AB} \cdot r_{AC} = (-3)(-3) + (-0.75)(1) + (1)(1.5) = 9.75$
We proceed to calculate the angle:
$\theta = \cos^{-1} \left(\dfrac{r_{AB} \cdot r_{AC}}{r_{AB} r_{AC}} \right) = \cos^{-1} \left(\dfrac{9.75}{(3.25)(3.50)}\right) = 31.0°$
