Animated Solution for Physics - Rotational Motion: A force $\mathbf{F}=4\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\hat{\mathbf{k}}$ is applied on an intersection point of $x=2$ plane and X-axis. The magnitude of torque of this force about a point $(2,3,4)$ is …… . (Round off to the nearest integer)

Calculating Torque in 3D

Torque is one of those beautiful physical quantities that forces us to think in three dimensions. It's not just about how hard you push, but where you push and from where you are observing that push. Let's break down this classic JEE problem step-by-step.

First, we need to figure out exactly where this force is acting. The question states that the force $\mathbf{F}=4\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\hat{\mathbf{k}}$ is applied at the intersection of the $x=2$ plane and the X-axis.

What does this mean geometrically? On the X-axis, the $y$ and $z$ coordinates are strictly zero. Since it intersects the $x=2$ plane, the $x$ coordinate must be $2$. Therefore, our point of application is $P(2,0,0)$. The position vector of this point from the origin is simply ${\mathbf{r}}_P=2\hat{\mathbf{i}}$.

Next, we identify the reference point—the point about which we are calculating the torque. The problem gives this as $Q(2,3,4)$. Its position vector is ${\mathbf{r}}_Q=2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\hat{\mathbf{k}}$.

Torque depends on the lever arm, which is the relative position vector pointing from the reference point to the point of application. Let's call this vector $\mathbf{r}$.

Substituting our vectors:

Now we invoke the master formula for torque: $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$.

A critical warning: The cross product is anti-commutative. $\mathbf{r}\times \mathbf{F}$ is NOT the same as $\mathbf{F}\times \mathbf{r}$. Always ensure the position vector $\mathbf{r}$ comes first. We will use the determinant method to evaluate this:

Let's expand this determinant carefully:

We have the torque vector, but the question asks for its magnitude. The magnitude of a 3D vector is the square root of the sum of the squares of its components.

The magnitude of the torque is exactly 20. This problem beautifully tests your ability to translate geometric constraints into vectors and execute a flawless cross product.