Help me confirm I got the right answer: finding the torque F

[maguss182]

Homework Statement

Can someone just go over this to see if I got it correct, I got this wrong in my midterm and I'm trying to redo it to make sure I got it correct the second time.

Relevant Equations

what tripped me up was getting the n unit vector, I got 1/sqrt40 originally, but I think I got it right this time.

 

[ergospherical]

I think you have the general idea, although it's not quite correct. Take any point on the line, say $\mathbf{r}_0 = (0,9,0)$, as you chose. The moment of $\mathbf{F}$ about this point is
\begin{align*}
\mathbf{G} = (\mathbf{r} - \mathbf{r}_0) \times \mathbf{F} = (6, -9, -1) \times (2,0,6) = (-54, -38, 18)
\end{align*}The unit vector along the line, as you wrote, is $\mathbf{n} = \frac{1}{\sqrt{58}} (7,0,-3)$. The moment of $\mathbf{F}$ about the line is the projection of $\mathbf{G}$ onto $\mathbf{n}$, which is a scalar:
\begin{align*}
\mathbf{G} \cdot \mathbf{n} = \frac{1}{\sqrt{58}} (-54, -38, 18) \cdot (7,0,-3) = \dots
\end{align*}(It doesn't matter which point on the line you choose, because if you consider the point, say, $\mathbf{r}_0 + \alpha \mathbf{n}$, then you have an extra term $-\alpha \mathbf{n} \times \mathbf{F}$ in the moment which vanishes upon taking the scalar product with $\mathbf{n}$.)

 

[maguss182]

oh shoot, so it would be 1/sqrt58 (-378, 0,-54)?

 

[ergospherical]

oh shoot, so it would be 1/sqrt58 (-378, 0,-54)?

Not quite, because the result of a dot product is a number, i.e. $(a_1,a_2,a_3) \cdot (b_1,b_2,b_3) = a_1 b_1 + a_2 b_2 + a_3 b_3$.

 

[maguss182]

right got it, thanks!