In your given exercise, let force $\vec F$ be $\begin{bmatrix} 7 \\ -2 \\0\end{bmatrix},$ its application point $A$ be $(2,2,0),$ and $B$ be the point $(7,-2,0).$

1. As $\vec{OA}$ and $\vec{OB}$ are position vectors, they are “bound vectors”.
2. Now, the torque $\vec\tau$ about point $O$ due to force $\vec F$ applied at point $A$ is a function of (depends on) the distance between $O$ and $\vec F.$ So, computing $\vec\tau$ intrinsically forbids $\vec F$ from freely translating (this will change said distance from $\left|\vec{OA}\right|\sin\theta$ to something else) but allows $\vec F$ to freely slide along its line of action (this does not affect said distance); in other words, for the purpose of computing $\vec\tau,\;\vec F$ is a “sliding vector”.
3. Finally, let's geometrically determine the sum of $\vec F$ and $\vec{OA}.$ We do this by applying the Parallelogram Law of Vector Addition, whose algorithm demands that the two vectors be freely translated (but not rotated, for that would alter them). Here, clearly $\vec F$ and $\vec{OA}$ are “free vectors”.

Notice that even within a single physical setup, the vector $$\vec F=\begin{bmatrix} 7 \\ -2 \\0\end{bmatrix}=\vec{OB}$$ is variously “free”, “sliding” and “bound”, depending on the context; by “context”, I mean what we are doing with the vector, in other words, what definition or theorem we are invoking for the task at hand.

The moral is this: after the physical scenario has been mathematically modelled (abstracted), the classification system <free vector versus sliding vector versus bound vector> is unnecessary; carefully studying the phrasing and conditions of definitions/theorems is all that's required to perform valid steps.

In your given exercise, let force $\vec F$ be $\begin{bmatrix} 7 \\ -2 \\0\end{bmatrix},$ its application point $A$ be $(2,2,0),$ and $B$ be the point $(7,-2,0).$

1. As $\vec{OA}$ and $\vec{OB}$ are position vectors, they are “bound vectors”.
2. Now, the torque $\vec\tau$ about point $O$ due to force $\vec F$ applied at point $A$ is a function of (depends on) the distance between $O$ and $\vec F.$ So, computing $\vec\tau$ intrinsically forbids $\vec F$ from freely translating (this will change said distance from $\left|\vec{OA}\right|\sin\theta$ to something else) but allows $\vec F$ to freely slide along its line of action (this does not affect said distance); in other words, for the purpose of computing $\vec\tau,\;\vec F$ is a “sliding vector”.
3. Finally, let's geometrically determine the sum of $\vec F$ and $\vec{OA}.$ We do this by applying the Parallelogram Law of Vector Addition, whose algorithm demands that the two vectors be freely translated (but not rotated, for that would alter them). Here, clearly $\vec F$ and $\vec{OA}$ are “free vectors”.

Notice that even within a single physical setup, the vector $$\vec F=\begin{bmatrix} 7 \\ -2 \\0\end{bmatrix}=\vec{OB}$$ is variously “free”, “sliding” and “bound”, depending on the context; by “context”, I mean what we are doing with the vector, in other words, what definition or theorem we are invoking for the task at hand.

The moral is this: after the physical scenario has been mathematically modelled (abstracted), the classification system <free vector versus sliding vector versus bound vector> is unnecessary; instead, carefully respecting the phrasing and conditions of definitions and theorems ensures that steps performed are valid.

In your given exercise, let force $\vec F$ be $\begin{bmatrix} 7 \\ -2 \\0\end{bmatrix},$ its application point $A$ be $(2,2,0),$ and $B$ be the point $(7,-2,0).$

1. As $\vec{OA}$ and $\vec{OB}$ are position vectors, they are “bound vectors”.
2. Now, the torque $\vec\tau$ about point $O$ due to force $\vec F$ applied at point $A$ is a function of (depends on) the distance between $O$ and $\vec F.$ So, computing $\vec\tau$ intrinsically forbids $\vec F$ from freely translating (this will change said distance from $\left|\vec{OA}\right|\sin\theta$ to something else) but allows $\vec F$ to freely slide along its line of action (this does not affect said distance); in other words, for the purpose of computing $\vec\tau,\;\vec F$ is a “sliding vector”.
3. Finally, let's geometrically determine the sum of $\vec F$ and $\vec{OA}.$ We do this by applying the Parallelogram Law of Vector Addition, whose algorithm demands that the two vectors be freely translated (but not rotated, for that would alter them). Here, clearly $\vec F$ and $\vec{OA}$ are “free vectors”.

Notice that even within a single physical setup, the vector $$\vec F=\begin{bmatrix} 7 \\ -2 \\0\end{bmatrix}=\vec{OB}$$ is variously “free”, “sliding” and “bound”, depending on the context; by “context”, I mean what we are doing with the vector, in other words, what definition or theorem we are invoking for the task at hand.

The moral is this: the classification system <free vector versus sliding vector versus bound vector> is not terribly relevant after the physical scenario has been mathematically modelled (abstracted).

In your given exercise, let force $\vec F$ be $\begin{bmatrix} 7 \\ -2 \\0\end{bmatrix},$ its application point $A$ be $(2,2,0),$ and $B$ be the point $(7,-2,0).$

1. As $\vec{OA}$ and $\vec{OB}$ are position vectors, they are “bound vectors”.
2. Now, the torque $\vec\tau$ about point $O$ due to force $\vec F$ applied at point $A$ is a function of (depends on) the distance between $O$ and $\vec F.$ So, computing $\vec\tau$ intrinsically forbids $\vec F$ from freely translating (this will change said distance from $\left|\vec{OA}\right|\sin\theta$ to something else) but allows $\vec F$ to freely slide along its line of action (this does not affect said distance); in other words, for the purpose of computing $\vec\tau,\;\vec F$ is a “sliding vector”.
3. Finally, let's geometrically determine the sum of $\vec F$ and $\vec{OA}.$ We do this by applying the Parallelogram Law of Vector Addition, whose algorithm demands that the two vectors be freely translated (but not rotated, for that would alter them). Here, clearly $\vec F$ and $\vec{OA}$ are “free vectors”.

Notice that even within a single physical setup, the vector $$\vec F=\begin{bmatrix} 7 \\ -2 \\0\end{bmatrix}=\vec{OB}$$ is variously “free”, “sliding” and “bound”, depending on the context; by “context”, I mean what we are doing with the vector, in other words, what definition or theorem we are invoking for the task at hand.

The moral is this: after the physical scenario has been mathematically modelled (abstracted), the classification system <free vector versus sliding vector versus bound vector> is unnecessary; carefully studying the phrasing and conditions of definitions/theorems is all that's required to perform valid steps.