# Torque about a Line in 3D space

given: Force Vector $$F = 2i - 3j + k$$

force acts on the point $$(1,5,2)$$

line $$\frac{x}{2} = y = \frac{z}{-2}$$

Known: $$T = n * (r X F)$$

'n' is a unit vector in the direction of the given line

r is the position vector

vectors, so this is a dot and cross product respectively

Find: torque about the given line

I'm not used to dealing with lines in 3D space defined in this manner, so I'm a bit stumped on how to find a unit vector on the given line.

• How do you know which of two orientations to use for $n$? If you’ve only interested in the magnitude of the torque, then it doesn’t make any difference, but torque also has a direction. – amd Feb 15 '19 at 0:59

If your line is defined as $$y=x/2=-z/2$$, then moving one unit along $$y$$ must take you 2 in the $$x$$ direction and $$-2$$ in the $$z$$ direction. So, $$(2,1,-2)$$ is a vector parallel to the line. The magnitude of that vector is $$\sqrt{4 + 1 + 4} = 3$$, so a unit length vector in the same direction would be, $$\frac{1}{3} (2,1,-2)$$
• Since the line passes through the origin, if $(1/2,1,-1/2)$ is indeed its direction vector, then that point must lie on the line. However, $x/2=1/4$ and $y=1$, so that test fails. – amd Feb 15 '19 at 1:16