A force of 39 kg. wt is acting at a point p ( -4,2,5) in the direaction ` 12hati -4hatj -3hatk` . The moment of this force about a line through the origin having the direction of ` 2hati -2hatj + hatk` is

Options

A. 76 units
B. - 76 units
C. ` 42hati + 144 hatj -24 hatk`
D. none of these

Explanation

To solve the problem step by step, we will follow the outlined process in the video transcript.

### Step 1: Calculate the Force Vector
Given the force of 39 kg wt acting in the direction of the vector \( 12 \hat{i} - 4 \hat{j} - 3 \hat{k} \), we first need to find the unit vector in the direction of the force.

1. Calculate the magnitude of the direction vector:
\[ \text{Magnitude} = \sqrt{12^2 + (-4)^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13 \]

2. Find the unit vector:
\[ \text{Unit vector} = \frac{1}{13}(12 \hat{i} - 4 \hat{j} - 3 \hat{k}) = \left(\frac{12}{13} \hat{i} - \frac{4}{13} \hat{j} - \frac{3}{13} \hat{k}\right) \]

3. Calculate the force vector:
\[ \text{Force} = 39 \times \text{Unit vector} = 39 \left(\frac{12}{13} \hat{i} - \frac{4}{13} \hat{j} - \frac{3}{13} \hat{k}\right) = 36 \hat{i} - 12 \hat{j} - 9 \hat{k} \]

### Step 2: Determine the Position Vector
The position vector \( \mathbf{r} \) from the origin to point \( P(-4, 2, 5) \) is:
\[ \mathbf{r} = -4 \hat{i} + 2 \hat{j} + 5 \hat{k} \]

### Step 3: Calculate the Moment of the Force
The moment of the force \( \mathbf{M} \) about the origin is given by the cross product \( \mathbf{r} \times \mathbf{F} \).

1. Set up the determinant for the cross product:
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 2 & 5 \\ 36 & -12 & -9 \end{vmatrix} \]

2. Calculate the determinant:
\[ \mathbf{M} = \hat{i} \begin{vmatrix} 2 & 5 \\ -12 & -9 \end{vmatrix} - \hat{j} \begin{vmatrix} -4 & 5 \\ 36 & -9 \end{vmatrix} + \hat{k} \begin{vmatrix} -4 & 2 \\ 36 & -12 \end{vmatrix} \]

- For \( \hat{i} \):
\[ = 2 \cdot (-9) - 5 \cdot (-12) = -18 + 60 = 42 \]

- For \( \hat{j} \):
\[ = -(-4 \cdot -9 - 5 \cdot 36) = -36 - 180 = -216 \Rightarrow 216 \]

- For \( \hat{k} \):
\[ = -4 \cdot -12 - 2 \cdot 36 = 48 - 72 = -24 \]

3. Combine the results:
\[ \mathbf{M} = 42 \hat{i} + 216 \hat{j} - 24 \hat{k} \]

### Step 4: Calculate the Moment about the Given Line
The direction vector of the line is \( \mathbf{a} = 2 \hat{i} - 2 \hat{j} + \hat{k} \).

1. Calculate the magnitude of \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

2. Find the unit vector \( \mathbf{u} \):
\[ \mathbf{u} = \frac{1}{3}(2 \hat{i} - 2 \hat{j} + \hat{k}) \]

3. Calculate the moment about the line:
\[ \text{Moment about the line} = \frac{\mathbf{M} \cdot \mathbf{u}}{|\mathbf{a}|} \]
\[ \mathbf{M} \cdot \mathbf{u} = (42 \hat{i} + 216 \hat{j} - 24 \hat{k}) \cdot \left(\frac{1}{3}(2 \hat{i} - 2 \hat{j} + \hat{k})\right) \]
\[ = \frac{1}{3} \left(42 \cdot 2 + 216 \cdot (-2) + (-24) \cdot 1\right) = \frac{1}{3} \left(84 - 432 - 24\right) = \frac{1}{3} \left(-372\right) = -124 \]

### Final Answer
The moment of the force about the line is \( -124 \) units.

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