The position of a particle moving in space varies with time `t` according as `:-`
`x(t)=3 cosomegat`
`y(t)=3sinomegat`
`z(t)=3t-8`
where `omega` is a constant. Minimum distance of particle from origin is `:-`

Options

A. `1` unit
B. `2` unit
C. `3` unit
D. `4` unit

Explanation

To find the minimum distance of the particle from the origin, we start with the given position functions:

1. **Position Functions**:
- \( x(t) = 3 \cos(\omega t) \)
- \( y(t) = 3 \sin(\omega t) \)
- \( z(t) = 3t - 8 \)

2. **Distance from the Origin**:
The distance \( d \) of the particle from the origin can be expressed as:
\[ d^2 = x^2 + y^2 + z^2 \]
Substituting the expressions for \( x(t) \), \( y(t) \), and \( z(t) \):
\[ d^2 = (3 \cos(\omega t))^2 + (3 \sin(\omega t))^2 + (3t - 8)^2 \]
Simplifying this:
\[ d^2 = 9 \cos^2(\omega t) + 9 \sin^2(\omega t) + (3t - 8)^2 \]
Using the identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \):
\[ d^2 = 9 + (3t - 8)^2 \]

3. **Finding Minimum Distance**:
To find the minimum distance, we need to minimize \( d^2 \). We can differentiate \( d^2 \) with respect to \( t \):
\[ \frac{d(d^2)}{dt} = 0 \]
The expression for \( d^2 \) is:
\[ d^2 = 9 + (3t - 8)^2 \]
Differentiating:
\[ \frac{d(d^2)}{dt} = 2(3t - 8) \cdot 3 = 6(3t - 8) \]
Setting the derivative to zero to find critical points:
\[ 6(3t - 8) = 0 \]
Solving for \( t \):
\[ 3t - 8 = 0 \implies t = \frac{8}{3} \]

4. **Calculating Minimum Distance**:
Now we substitute \( t = \frac{8}{3} \) back into the expression for \( d^2 \):
\[ d^2 = 9 + (3 \cdot \frac{8}{3} - 8)^2 \]
Simplifying:
\[ d^2 = 9 + (8 - 8)^2 = 9 + 0 = 9 \]
Therefore, the minimum distance \( d \) is:
\[ d = \sqrt{9} = 3 \]

Thus, the minimum distance of the particle from the origin is **3 units**.