\(\vec{M_O}=\kNm{\left\langle 1.114, 0.911, -0.446 \right\rangle}\)
Start the problem by using the position information and tension magnitude to find the force vector \(\vec{F}_{BD}\text{.}\) This will be done in three steps:
Find the position vector \(\vec{BD}\): Find position vectors by either subtracting the start-point coordinates from the end-point coordinates or focusing on the changes in the position components from \(B\) to \(D\text{.}\)
\begin{align*}
\vec{BD} \amp = D-B\\
\amp=\m{(-0.9, 1.1, 0)} - \m{(0.4, 0, 1.0)} \\
\amp = \m{\left\langle 1.3, -1.1, 1 \right\rangle}
\end{align*}
Find the unit vector of \(\vec{BD}\): Compute a unit vector by dividing \(\vec{BD}\) by the total length of \(BD\text{.}\)
\begin{align*}
BD \amp = |\vec{BD}|=\sqrt{1.3^2+(-1.1)^2+1.0^2}\\
\amp = \m{1.975}\\
\vec{\widehat{BD}} \amp = \frac{\vec{BD}}{BD}\\
\amp = \frac {\m{\left\langle 1.3, -1.1, 1 \right\rangle}}{\m{1.975}}\\
\amp = \left\langle 0.658 -0.557 0.506 \right\rangle
\end{align*}
Note that \(\vec{\widehat{BD}}\) is unitless and is the pure direction of \(\vec{BD}\text{.}\)
Multiply the unit vector by force magnitude: Now multiply \(\vec{\widehat{BD}}\) by the \(\kN{2}\) force magnitude to find the force components.
\begin{align*}
\vec{F}_{BD} \amp = F_{BD}(\vec{\widehat{BD}})\\
\amp =\kN{2}\left\langle 0.658, -0.557, 0.506 \right\rangle\\
\amp = \kN{\left\langle 1.317, -1.114, 1.013 \right\rangle}
\end{align*}
Next, find the moment arm from point \(O\) to the line of action of the force. There are two obvious options for moment arms, either \(\vec{r}_{OB}\) or \(\vec{r}_{OB}\text{.}\) To demonstrate how both moment arms give the same answer, solutions for both moment arms will be shown.
Option 1: Moment using \(\vec{r}_{OB}\)
A thin plate \(OABC\) sits in the xy plane. Cable BD pulls with a tension of 2 kN down to anchor point D. A position vector r_OB goes from the origin to the tail of the 2 kN force vector.
Moment arm \(\vec{r}_{OB}\) starts at the point we are taking the moment around, \(O\text{,}\) and ends at the point \(B\text{.}\)
\begin{align*}
\vec{OB}\amp =B-O\\
\amp =\m{(-0.9, 1.1, 0)} - \m{(0, 0, 0)}\\
\amp = \m{\left\langle -0.9, 1.1, 0 \right\rangle}
\end{align*}
Cross \(\vec{r}_{OB}\) with \(\vec{F}_{BD}\) to find the moment of \(\vec{F}_{BD}\) about point \(O\text{.}\)
\begin{align*}
\vec{M}_O \amp = \vec{r}_{OB} \times \vec{F}_{BD}\\
\amp = \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ -0.9 \amp 1.1 \amp 0 \\ 1.317 \amp -1.114 \amp 1.013 \end{vmatrix}\\
\amp = \kN{\left\langle 1.114, 0.911, -0.446 \right\rangle}
\end{align*}
Option 2: Moment using \(\vec{r}_{OD}:\)
A thin plate \(OABC\) sits in the xy plane. Cable BD pulls with a tension of 2 kN down to anchor point D. A position vector r_OB goes from the origin to the tail of the 2 kN force vector.
Moment arm \(\vec{r}_{OD}\) starts at the point we are taking the moment around, \(O\text{,}\) and ends at the point \(D\text{.}\)
\begin{align*}
\vec{OD}\amp=D-O\\
\amp =\m{(0.4, 0, 1.0)} - \m{(0, 0, 0)} \\
\amp = \m{\left\langle 0.4, 0, 1.0 \right\rangle}
\end{align*}
Cross \(\vec{r}_{OD}\) with \(\vec{F}_{BD}\) to find the moment of \(\vec{F}_{BD}\) about point \(O\text{.}\)
\begin{align*}
\vec{M_O} \amp = \vec{r}_{OD} \times \vec{F}_{BD}\\
\amp = \begin{vmatrix} \ihat \amp \jhat \amp \khat \\ 0.4 \amp 0 \amp 1.0\\ 1.317 \amp -1.114 \amp 1.013 \end{vmatrix}\\
\amp = \kN{\left\langle 1.114, 0.911, -0.446 \right\rangle}
\end{align*}
It is worth your effort to compute moments both ways for this example, or another problem, to prove to yourself that the answers work out exactly the same with different moment arms. Technically, you could select a position vector from anywhere on line \(\vec{BD}\) and get the correct answer, but \(\vec{r}_{OB}\) or \(\vec{r}_{OB}\) are the only two between defined points in this problem.
Drawing \(\vec{M}_O\text{,}\) demonstrates that a moment vector direction is both 1) the axis of rotation caused by \(\vec{T}_{BD}\) around point \(O\text{,}\) with the moment aligning to your thumb and the moment rotating around your fingers from the right-hand rule and 2) that \(\vec{M}_O\) is perpendicular to the plane formed by \(\vec{T}_{BD}\) and \(\vec{T}_{BD}\text{.}\) Recall that all cross products result in vectors perpendicular to the two crossed vectors.
