Explanation:
Given that,
Force, [tex]F=((-8i)+6j)\ N[/tex]
Position of the particle, [tex]r=(3i+4j)\ m[/tex]
(a) The toque on a particle about the origin is given by :
[tex]\tau=F\times r[/tex]
[tex]\tau=((-8i)+6j) \times (3i+4j)[/tex]
Taking the cross product of above two vectors, we get the value of torque as :
[tex]\tau=(0+0-50k)\ N-m[/tex]
(b) Let [tex]\theta[/tex] is the angle between r and F. The angle between two vectors is given by :
[tex]cos\theta=\dfrac{r.F}{|r|.|F|}[/tex]
[tex]cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }[/tex]
[tex]cos\theta=\dfrac{0}{50}[/tex]
[tex]\theta=90^{\circ}[/tex]
