Example 3.92. Cross Product Using Components of Vectors.
A particle located at \((x, y, z) = (3, -2, 5) \text{,}\) where distances are in the unit \(\text{m} \text{,}\) is subject to a force whose components are \((F_x, F_y, F_z) = \)\((-10, 30, -20 )\) in units of \(\text{N}\text{.}\) Find the components of torque on the particle given by \(\vec r \times \vec F\text{.}\)
Answer.
\(-110\text{ N.m }\hat i + 10\text{ N.m }\hat j + 70\text{ N.m }\hat k.\)
Solution.
You may recall that \(x, \, y,\, z \) are the components of the position vector \(\vec r \text{.}\) Therefore to get the components of \(\vec r \times \vec F\) we just use the formulas given.
\begin{align*}
(\vec r \times \vec F)_x \amp = y F_z - z F_y \\
\amp = (-2)\times(-20) - 5\times 30 = -110
\end{align*}
\begin{align*}
(\vec r \times \vec F)_y \amp = z F_x - x F_z \\
\amp = 5\times(-10) - 3\times (-20) = 10
\end{align*}
\begin{align*}
(\vec r \times \vec F)_z \amp = x F_y - y F_x \\
\amp = 3\times 30 - (-2)\times (-10) = 70
\end{align*}
Therefore,
\begin{equation*}
\vec \tau = -110\text{ N.m }\hat i + 10\text{ N.m }\hat j + 70\text{ N.m }\hat k.
\end{equation*}
