A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector `vecr=(2.00m)hati-(3.00m)hatj+(2.00m)hatk`, the force is given by `vecF=F_(x)hati+(7.00N)hatj-(6.00N)hatk` and the corresponding torque about the origin is `vectau=(400N*m)hati+(2.00N*m)hatj-(1.00N*m)hatk`. Determine `F_(x)`.

Explanation

To solve the problem, we need to find the value of \( F_x \) given the position vector \( \vec{r} \), the force vector \( \vec{F} \), and the torque vector \( \vec{\tau} \). The relationship between torque, position vector, and force vector is given by the equation:

\[ \vec{\tau} = \vec{r} \times \vec{F} \]

### Step 1: Write down the vectors
The position vector \( \vec{r} \) and the force vector \( \vec{F} \) are given as follows:
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