physicsvectors
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.
A single force can be represented by an arrow (a simple assumption)
An arrow is one-dimensional.
As by Google, a
Vector is:
a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another.
Quantity is:
the amount or number of a material or abstract thing not usually estimated by spatial measurement.
We know that the length of an arrow represents the power of a force, or in other words, the quantity. Also, we know that the direction of an arrow is it's, well direction, and the magnitude being the quantity. And, also, it is actually determining the position of a point in space relative to another! The comparison is, well an origin on a set of coordinates, longitude and latitude in real life, and there are also many more examples!
From this, we conclude that it is not necessary to classify as something else.
As for your last question, I don't understand what is meant by such quantities, so therefore, I cannot answer that (probably because of my poor comprehension...)
Imagine this line $r=a+\lambda\vec{d}$ is intersecting plane $\pi: \vec{r}.\hat n = k$
The cross product of $\hat n \times \hat{d} = \hat{u_1}$ this $\hat{u_1}$ will be the unit vector normal to the plane of line containing line $r=a+\lambda\vec{d}$ and plane $\pi: \vec{r}.\hat n = k$
Now, we take the cross product of $\hat{u_1}$ and $\hat n$: $\hat {u_2} = \hat {u_1} \times \hat n$ will be the required unit vector.
Note: Here the $\hat {u_2}$ depends on the unit vector $\hat d$ I mean $\hat u_2 = \hat {u_1} \times \hat n $ or $\hat {u_2} = \hat n \times \hat {u_1}$
Find the direction vector of line PQ.
Here, I used unit vectors only as you were interested in the direction of $\vec{PQ}$;
Best Answer
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, $$ \frac{1}{3} (2,1,-2) $$