Find the moment about the point (3, 4, 5) of the force through the point `(1, 2, -3)` having components equal to `-2, 3, -4`. What is the moment of the same force about the line through the origin having direction - ratios `lt 4, -2, 5 gt` ?

Answer

To solve the problem of finding the moment about the point (3, 4, 5) of the force through the point (1, 2, -3) with components (-2, 3, -4), and then finding the moment of the same force about the line through the origin with direction ratios (4, -2, 5), we can follow these steps:

### Step 1: Define the Given Information
- Point A (where we want to find the moment): \( A(3, 4, 5) \)
- Point B (where the force is applied): \( B(1, 2, -3) \)
- Force vector \( \mathbf{F} = -2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k} \)

### Step 2: Calculate the Position Vector \( \mathbf{r_{AB}} \)
The position vector from point A to point B is given by:
\[ \mathbf{r_{AB}} = \mathbf{r_B} - \mathbf{r_A} = (1 - 3)\mathbf{i} + (2 - 4)\mathbf{j} + (-3 - 5)\mathbf{k} \]
\[ \mathbf{r_{AB}} = -2\mathbf{i} - 2\mathbf{j} - 8\mathbf{k} \]

### Step 3: Calculate the Moment \( \mathbf{M_A} \)
The moment about point A is given by the cross product of the position vector \( \mathbf{r_{AB}} \) and the force vector \( \mathbf{F} \):
\[ \mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F} \]
Using the determinant form:
\[ \mathbf{M_A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & -2 & -8 \\ -2 & 3 & -4 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{M_A} = \mathbf{i}((-2)(-4) - (-8)(3)) - \mathbf{j}((-2)(-4) - (-8)(-2)) + \mathbf{k}((-2)(3) - (-2)(-2)) \]
\[ = \mathbf{i}(8 + 24) - \mathbf{j}(8 - 16) + \mathbf{k}(-6 + 4) \]
\[ = 32\mathbf{i} + 8\mathbf{j} - 2\mathbf{k} \]

### Step 4: Calculate the Moment about the Line
To find the moment about the line with direction ratios (4, -2, 5), we first need the unit vector in the direction of the line:
\[ \mathbf{b} = 4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} \]
The magnitude of \( \mathbf{b} \):
\[ |\mathbf{b}| = \sqrt{4^2 + (-2)^2 + 5^2} = \sqrt{16 + 4 + 25} = \sqrt{45} \]
The unit vector \( \mathbf{b_{unit}} \):
\[ \mathbf{b_{unit}} = \frac{1}{\sqrt{45}}(4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}) \]

### Step 5: Calculate the Moment about the Line
The moment about the line is given by:
\[ \text{Moment about line} = \frac{\mathbf{M_A} \cdot \mathbf{b}}{|\mathbf{b}|} \]
Calculating the dot product:
\[ \mathbf{M_A} \cdot \mathbf{b} = (32\mathbf{i} + 8\mathbf{j} - 2\mathbf{k}) \cdot (4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}) \]
\[ = 32 \cdot 4 + 8 \cdot (-2) - 2 \cdot 5 = 128 - 16 - 10 = 102 \]
Thus, the moment about the line is:
\[ \text{Moment about line} = \frac{102}{\sqrt{45}} \]

### Final Answers
1. Moment about point A: \( \mathbf{M_A} = 32\mathbf{i} + 8\mathbf{j} - 2\mathbf{k} \)
2. Moment about the line: \( \frac{102}{\sqrt{45}} \)