Answer
(a) $\vec A \cdot \vec B = 6.0$
(b) $\phi= 82.1^o$
Work Step by Step
(a)
$$\vec A \cdot \vec B = A_xB_x + A_yB_y = (4.00)(5.00) + (7.00)(-2.00)$$ $$\vec A \cdot \vec B = 20.0 - 14.0 = 6.0$$
(b)
Solving Eq. 1.16 for cos $\phi$:
$$\vec A \cdot \vec B = ABcos \phi \longrightarrow cos \phi = \frac{\vec A \cdot \vec B}{AB}$$
- Calculate A and B:
$$A = \sqrt {A_x^2 + A_y^2} = \sqrt{ 4.00^2 + 7.00^2} = 8.06$$ $$B = \sqrt{B_x^2 + A_x^2} = \sqrt{5.00^2 + (-2.00)^2} = 5.39$$
$$cos \phi = \frac{6.0 }{(8.06)(5.39)} = 0.138$$ $$\phi = arccos(0.138) = 82.1^o$$
** Since both components are positive, the angle is correct.
