A particle moves with a velocity `v=(5hati-3hatj + 6hatk) ms^(-1)` under the influence of a constant force
`F =(10hati + 10hatj + 20hath) N`, the instantaneous power applied to the particle is.

Options

A. 200 `Js^(-1)`
B. 40 `Js^(-1)`
C. 140 `Js^(-1)`
D. 170 `Js^(-1)`

Explanation

To find the instantaneous power applied to the particle, we will use the formula for instantaneous power, which is given by the dot product of the force vector and the velocity vector.

### Step-by-Step Solution:

1. **Identify the given vectors**:
- Velocity vector \( \mathbf{v} = 5 \hat{i} - 3 \hat{j} + 6 \hat{k} \) m/s
- Force vector \( \mathbf{F} = 10 \hat{i} + 10 \hat{j} + 20 \hat{k} \) N

2. **Write the formula for instantaneous power**:
\[ P_{\text{instantaneous}} = \mathbf{F} \cdot \mathbf{v} \]
where \( \cdot \) denotes the dot product.

3. **Calculate the dot product**:
The dot product of two vectors \( \mathbf{A} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \mathbf{B} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \) is calculated as:
\[ \mathbf{A} \cdot \mathbf{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]

For our vectors:
- \( a_1 = 10, a_2 = 10, a_3 = 20 \) (components of \( \mathbf{F} \))
- \( b_1 = 5, b_2 = -3, b_3 = 6 \) (components of \( \mathbf{v} \))

Now, substituting these values into the dot product formula:
\[ P_{\text{instantaneous}} = (10)(5) + (10)(-3) + (20)(6) \]

4. **Perform the calculations**:
- \( (10)(5) = 50 \)
- \( (10)(-3) = -30 \)
- \( (20)(6) = 120 \)

Now, adding these results together:
\[ P_{\text{instantaneous}} = 50 - 30 + 120 \]
\[ P_{\text{instantaneous}} = 140 \text{ Joules per second} \]

5. **Final answer**:
The instantaneous power applied to the particle is \( 140 \) Joules per second.