Animated Solution for Physics - Rotational Motion: A force $\mathbf{F}=(\hat{i}+2\hat{j}+3\hat{k})$ N acts at a point $(4\hat{i}+3\hat{j}-\hat{k})$ m. Then, the magnitude of torque about the point $(\hat{i}+2\hat{j}+\hat{k})$ m will be $\sqrt{x}$ N-m. The value of $x$ is …… .

Imagine you are trying to open a heavy door. The effectiveness of your push doesn't just depend on how hard you push, but also on where you push it relative to the hinges. This rotational effectiveness is what we call Torque.

In this problem, we are given a force $\mathbf{F}=\hat{i}+2\hat{j}+3\hat{k}$ acting at a specific point in space, ${\mathbf{r}}_1=4\hat{i}+3\hat{j}-\hat{k}$. However, we want to find the torque about a completely different point, ${\mathbf{r}}_2=\hat{i}+2\hat{j}+\hat{k}$.

To find the torque, we first need the relative position vector, $\mathbf{r}$. This vector acts as our "lever arm" and points from the axis of rotation (${\mathbf{r}}_2$) to the point where the force is applied (${\mathbf{r}}_1$).

Substituting the given coordinates:

Now that we have our lever arm $\mathbf{r}$ and our force $\mathbf{F}$, we can calculate the torque $\tau$. Torque is defined as the cross product of these two vectors:

To compute this, we set up a determinant. I know this determinant looks terrifying, but let's take a breath and expand it systematically.

Expanding along the first row:

We have the torque vector, but the question asks for its magnitude, $|\tau |$. The magnitude of a 3D vector is found using the 3D Pythagorean theorem:

The problem states that the magnitude is $\sqrt{x}$. By comparing our result with the given expression, we can confidently conclude that: