A particle is displaced from `A=(2,2,4)` to `B=(5, -3,-1)` . A constant force of `34N` acts in the direction of `vec(AP)`. Where `P=(10,2,-11)`. (Coordinates are in m).
(i) Find the `(vec(F))` .
(ii) Find the work done by the force to cause a displacement.

Explanation

To solve the problem step by step, we will find the force vector \( \vec{F} \) and then calculate the work done by this force during the displacement from point A to point B.

### Step 1: Find the vector \( \vec{AP} \)

The vector \( \vec{AP} \) is calculated using the coordinates of points \( P \) and \( A \).

\[ \vec{AP} = \vec{P} - \vec{A} = (10, 2, -11) - (2, 2, 4) \]

Calculating the components:

\[ \vec{AP} = (10 - 2, 2 - 2, -11 - 4) = (8, 0, -15) \]

### Step 2: Find the magnitude of \( \vec{AP} \)

The magnitude of \( \vec{AP} \) is given by:

\[ |\vec{AP}| = \sqrt{(8)^2 + (0)^2 + (-15)^2} = \sqrt{64 + 0 + 225} = \sqrt{289} = 17 \]

### Step 3: Find the unit vector in the direction of \( \vec{AP} \)

The unit vector \( \hat{u}_{AP} \) in the direction of \( \vec{AP} \) is:

\[ \hat{u}_{AP} = \frac{\vec{AP}}{|\vec{AP}|} = \left(\frac{8}{17}, 0, \frac{-15}{17}\right) \]

### Step 4: Find the force vector \( \vec{F} \)

Given that the magnitude of the force \( |\vec{F}| = 34 \, \text{N} \), we can find the force vector \( \vec{F} \):

\[ \vec{F} = |\vec{F}| \cdot \hat{u}_{AP} = 34 \cdot \left(\frac{8}{17}, 0, \frac{-15}{17}\right) \]

Calculating the components:

\[ \vec{F} = \left(34 \cdot \frac{8}{17}, 34 \cdot 0, 34 \cdot \frac{-15}{17}\right) = (16, 0, -30) \, \text{N} \]

### Step 5: Find the displacement vector \( \vec{AB} \)

The displacement vector \( \vec{AB} \) is calculated as:

\[ \vec{AB} = \vec{B} - \vec{A} = (5, -3, -1) - (2, 2, 4) \]

Calculating the components:

\[ \vec{AB} = (5 - 2, -3 - 2, -1 - 4) = (3, -5, -5) \]

### Step 6: Calculate the work done by the force

The work done \( W \) by the force is given by the dot product of the force vector and the displacement vector:

\[ W = \vec{F} \cdot \vec{AB} \]

Calculating the dot product:

\[ W = (16, 0, -30) \cdot (3, -5, -5) = (16 \cdot 3) + (0 \cdot -5) + (-30 \cdot -5) \]

Calculating each term:

\[ W = 48 + 0 + 150 = 198 \, \text{J} \]

### Final Answers

(i) The force vector \( \vec{F} = (16, 0, -30) \, \text{N} \)

(ii) The work done by the force is \( W = 198 \, \text{J} \)

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