Cross Product — Question 5

Question 5 – Solution

Keep the component order and signs organized when expanding the determinant. After computing the cross product, use a dot-product check or the relevant magnitude formula to interpret it.

Torque is $\vec\tau=\vec r\times\vec F$.

Compute $\vec\tau=\langle0.10(-20),\,-0.30(-20),\,0.30(40)\rangle=\langle-2,6,12\rangle$ N$\cdot$m.

Its magnitude is $\sqrt{4+36+144}=\boxed{2\sqrt{46}\text{ N}\cdot\text{m}}$.

The result follows from the defining vector formulas used above, and each component, magnitude, or scalar condition has been checked against the information in the question.

The cross product produces a vector perpendicular to both inputs. Its direction follows the right-hand rule, while its magnitude records the area of the spanned parallelogram.