The moment of the force `5i+10j+16k` acting at the point P, `2i-7j+10k` about the point O, `-5i+6j-10k` is

Options

A. `20i-12j+135k`
B. `-408i-12j+135k`
C. `20k`
D. none

Answer

To find the moment of the force \( \mathbf{F} = 5\mathbf{i} + 10\mathbf{j} + 16\mathbf{k} \) acting at the point \( P = 2\mathbf{i} - 7\mathbf{j} + 10\mathbf{k} \) about the point \( O = -5\mathbf{i} + 6\mathbf{j} - 10\mathbf{k} \), we will follow these steps:

### Step 1: Calculate the position vector \( \mathbf{R} \)
The position vector \( \mathbf{R} \) from point \( O \) to point \( P \) is given by:
\[ \mathbf{R} = \mathbf{P} - \mathbf{O} \]
Substituting the values:
\[ \mathbf{R} = (2\mathbf{i} - 7\mathbf{j} + 10\mathbf{k}) - (-5\mathbf{i} + 6\mathbf{j} - 10\mathbf{k}) \]
Calculating this gives:
\[ \mathbf{R} = 2\mathbf{i} - 7\mathbf{j} + 10\mathbf{k} + 5\mathbf{i} - 6\mathbf{j} + 10\mathbf{k} \]
\[ \mathbf{R} = (2 + 5)\mathbf{i} + (-7 - 6)\mathbf{j} + (10 + 10)\mathbf{k} \]
\[ \mathbf{R} = 7\mathbf{i} - 13\mathbf{j} + 20\mathbf{k} \]

### Step 2: Use the moment formula
The moment \( \mathbf{M} \) of the force about point \( O \) is given by the cross product:
\[ \mathbf{M} = \mathbf{R} \times \mathbf{F} \]
Substituting the vectors:
\[ \mathbf{M} = (7\mathbf{i} - 13\mathbf{j} + 20\mathbf{k}) \times (5\mathbf{i} + 10\mathbf{j} + 16\mathbf{k}) \]

### Step 3: Set up the determinant for the cross product
We will use the determinant to compute the cross product:
\[ \mathbf{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 7 & -13 & 20 \\ 5 & 10 & 16 \end{vmatrix} \]

### Step 4: Calculate the determinant
Expanding the determinant:
\[ \mathbf{M} = \mathbf{i} \begin{vmatrix} -13 & 20 \\ 10 & 16 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 7 & 20 \\ 5 & 16 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 7 & -13 \\ 5 & 10 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \( \mathbf{i} \):
\[ \begin{vmatrix} -13 & 20 \\ 10 & 16 \end{vmatrix} = (-13)(16) - (20)(10) = -208 - 200 = -408 \]

2. For \( \mathbf{j} \):
\[ \begin{vmatrix} 7 & 20 \\ 5 & 16 \end{vmatrix} = (7)(16) - (20)(5) = 112 - 100 = 12 \]

3. For \( \mathbf{k} \):
\[ \begin{vmatrix} 7 & -13 \\ 5 & 10 \end{vmatrix} = (7)(10) - (-13)(5) = 70 + 65 = 135 \]

### Step 5: Combine the results
Putting it all together:
\[ \mathbf{M} = -408\mathbf{i} - 12\mathbf{j} + 135\mathbf{k} \]

### Final Answer
Thus, the moment of the force about point \( O \) is:
\[ \mathbf{M} = -408\mathbf{i} - 12\mathbf{j} + 135\mathbf{k} \]