Let F be the force acting on a particle having position vector r and τ be the torque of this force a

Q: Why is torque defined as a cross product?

Torque measures the effectiveness of a force in producing rotation. Only the component of force perpendicular to the position vector causes rotation. The cross product naturally extracts this perpendicular component and assigns a direction to the axis of rotation using the right-hand rule.

Q: Can $\mathbf{r} \cdot \boldsymbol{\tau}$ be non-zero if the force is not perpendicular to $\mathbf{r}$?

No. Even if the angle between $\mathbf{r}$ and $\mathbf{F}$ is not $90^\circ$, the resulting cross product $\boldsymbol{\tau}$ will still be strictly perpendicular to the plane containing both $\mathbf{r}$ and $\mathbf{F}$. Therefore, $\boldsymbol{\tau}$ will always be perpendicular to $\mathbf{r}$, making their dot product zero.

Q: What does the dot product physically represent here?

The dot product $\mathbf{A} \cdot \mathbf{B}$ represents the projection of vector $\mathbf{A}$ along the direction of vector $\mathbf{B}$. Since torque is the axis of rotation, and $\mathbf{r}$ and $\mathbf{F}$ lie in the plane of rotation, they have zero projection along the axis of rotation.

Q: Is this property true for angular momentum as well?

Yes! Angular momentum is defined as $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. By the exact same geometric logic, $\mathbf{L}$ is perpendicular to both $\mathbf{r}$ and $\mathbf{p}$, meaning $\mathbf{r} \cdot \mathbf{L} = 0$ and $\mathbf{p} \cdot \mathbf{L} = 0$.

Q: What happens if the force is parallel to the position vector?

If $\mathbf{F}$ is parallel (or anti-parallel) to $\mathbf{r}$, the angle between them is $0^\circ$ (or $180^\circ$). The cross product $\mathbf{r} \times \mathbf{F}$ will be the zero vector ($\boldsymbol{\tau} = \mathbf{0}$). In this trivial case, the dot products $\mathbf{r} \cdot \mathbf{0}$ and $\mathbf{F} \cdot \mathbf{0}$ are still zero.

Question

Let $\mathbf{F}$ be the force acting on a particle having position vector $\mathbf{r}$ and $\boldsymbol{	au}$ be the torque of this force about the origin. Then,

Options:

(A) $\mathbf{r} \cdot \boldsymbol{	au} = 0$ and $\mathbf{F} \cdot \boldsymbol{	au} 
eq 0$

(B) $\mathbf{r} \cdot \boldsymbol{	au} 
eq 0$ and $\mathbf{F} \cdot \boldsymbol{	au} = 0$

(C) $\mathbf{r} \cdot \boldsymbol{	au} 
eq 0$ and $\mathbf{F} \cdot \boldsymbol{	au} 
eq 0$

(D) $\mathbf{r} \cdot \boldsymbol{	au} = 0$ and $\mathbf{F} \cdot \boldsymbol{	au} = 0$

Sigma Percentile · Accepted Answer

This problem is a beautiful test of your understanding of vector mathematics, specifically the geometric interpretations of the cross product and the dot product. Let's break down the physical and mathematical realities of torque.

### The Mathematical Definition of Torque

In rotational mechanics, torque ($\boldsymbol{	au}$) is the rotational equivalent of linear force. It measures the tendency of a force to rotate an object about an axis or pivot. Mathematically, it is defined as the cross product of the position vector ($\mathbf{r}$) and the force vector ($\mathbf{F}$):

$$\boldsymbol{	au} = \mathbf{r} 	imes \mathbf{F}$$

Here, $\mathbf{r}$ is the vector pointing from the origin (or pivot point) to the point where the force is applied.

### The Geometry of the Cross Product

The cross product of two vectors produces a third vector that has a very specific geometric property: **it is always strictly perpendicular (orthogonal) to the plane that contains the original two vectors.**

Imagine placing the vectors $\mathbf{r}$ and $\mathbf{F}$ flat on a table. The cross product $\boldsymbol{	au}$ will point straight up towards the ceiling (or straight down through the floor, depending on the right-hand rule). Because $\boldsymbol{	au}$ is perpendicular to the entire table, it must be perpendicular to every line drawn on that table.

Therefore, the torque vector $\boldsymbol{	au}$ is perpendicular to the position vector $\mathbf{r}$, and it is also perpendicular to the force vector $\mathbf{F}$.

### The Dot Product and Orthogonality

Now, let's look at the dot product. The dot product of any two vectors $\mathbf{A}$ and $\mathbf{B}$ is defined as:

$$\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(	heta)$$

where $	heta$ is the angle between them.

We just established that the angle between $\mathbf{r}$ and $\boldsymbol{	au}$ is exactly $90^\circ$. Similarly, the angle between $\mathbf{F}$ and $\boldsymbol{	au}$ is exactly $90^\circ$. Since $\cos(90^\circ) = 0$, the dot product of any two perpendicular vectors is always zero.

Consequently, we can definitively state:

$$\mathbf{r} \cdot \boldsymbol{	au} = 0$$
$$\mathbf{F} \cdot \boldsymbol{	au} = 0$$

This confirms that option (d) is the mathematically rigorous and correct answer. It is a fundamental property of the cross product that $\mathbf{A} \cdot (\mathbf{A} 	imes \mathbf{B}) = 0$ and $\mathbf{B} \cdot (\mathbf{A} 	imes \mathbf{B}) = 0$.

Animated Solution for Physics - Rotational Motion: Let $F$ be the force acting on a particle having position vector $r$ and $τ$ be the torque of this force about the origin. Then,

Select Answer:

Visualized Solution

$τ = r \times F$

$τ ⊥ r and τ ⊥ F$

$r \cdot τ = 0 and F \cdot τ = 0$

The Sigma Insight: Torque and Angular Momentum

The Mathematical Definition of Torque

The Geometry of the Cross Product

The Dot Product and Orthogonality

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Animated Solution for Physics - Rotational Motion: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let F be the force acting on a particle having position vector r and τ be the torque of this force about the origin. Then,

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τ=r×F

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τ⊥r and τ⊥F

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r⋅τ=0 and F⋅τ=0