The intensity of the resulting force at point A is:

To find the intensity (magnitude) of the resulting force at point \( A \), we first need to determine the components of the forces \( F_{AB} \) and \( F_{AC} \) in the \( x \), \( y \), and \( z \) directions. Then, we can sum these components to find the resultant force vector and its magnitude.

### Step 1: Determine the Unit Vectors for \( \mathbf{F_{AB}} \) and \( \mathbf{F_{AC}} \)

#### 1.1 Vector \( \mathbf{AB} \)
The position vector from \( A \) to \( B \) is:
\[
\overrightarrow{AB} = (2 - 0) \mathbf{i} + (0 - 4) \mathbf{j} + (0 - 4) \mathbf{k} = 2\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}
\]
The magnitude of \( \overrightarrow{AB} \) is:
\[
|\overrightarrow{AB}| = \sqrt{(2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 \, \text{m}
\]
The unit vector \( \mathbf{u_{AB}} \) is:
\[
\mathbf{u_{AB}} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|} = \frac{1}{6}(2\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}) = \frac{1}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} - \frac{2}{3}\mathbf{k}
\]

#### 1.2 Vector \( \mathbf{AC} \)
The position vector from \( A \) to \( C \) is:
\[
\overrightarrow{AC} = (4 - 0) \mathbf{i} + (0 - 4) \mathbf{j} + (0 - 4) \mathbf{k} = 4\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}
\]
The magnitude of \( \overrightarrow{AC} \) is:
\[
|\overrightarrow{AC}| = \sqrt{(4)^2 + (-4)^2 + (-4)^2} = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3} \, \text{m}
\]
The unit vector \( \mathbf{u_{AC}} \) is:
\[
\mathbf{u_{AC}} = \frac{\overrightarrow{AC}}{|\overrightarrow{AC}|} = \frac{1}{4\sqrt{3}}(4\mathbf{i} - 4\mathbf{j} - 4\mathbf{k}) = \frac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} - \mathbf{k})
\]

### Step 2: Determine the Force Vectors \( \mathbf{F_{AB}} \) and \( \mathbf{F_{AC}} \)

\[
\mathbf{F_{AB}} = F_{AB} \cdot \mathbf{u_{AB}} = 100 \cdot \left( \frac{1}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} - \frac{2}{3}\mathbf{k} \right) = \left(\frac{100}{3}\right)\mathbf{i} - \left(\frac{200}{3}\right)\mathbf{j} - \left(\frac{200}{3}\right)\mathbf{k}
\]
\[
\mathbf{F_{AC}} = F_{AC} \cdot \mathbf{u_{AC}} = 120 \cdot \frac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} - \mathbf{k}) = \left(\frac{120}{\sqrt{3}}\right)(\mathbf{i} - \mathbf{j} - \mathbf{k}) = 40\sqrt{3}\mathbf{i} - 40\sqrt{3}\mathbf{j} - 40\sqrt{3}\mathbf{k}
\]

### Step 3: Sum the Force Vectors to Find the Resultant Force \( \mathbf{F_R} \)

\[
\mathbf{F_R} = \mathbf{F_{AB}} + \mathbf{F_{AC}} 
\]
\[
\mathbf{F_R} = \left(\frac{100}{3} + 40\sqrt{3}\right)\mathbf{i} + \left(-\frac{200}{3} - 40\sqrt{3}\right)\mathbf{j} + \left(-\frac{200}{3} - 40\sqrt{3}\right)\mathbf{k}
\]

### Step 4: Calculate the Magnitude of the Resultant Force

The magnitude of the resultant force is given by:
\[
|\mathbf{F_R}| = \sqrt{\left(\frac{100}{3} + 40\sqrt{3}\right)^2 + \left(-\frac{200}{3} - 40\sqrt{3}\right)^2 + \left(-\frac{200}{3} - 40\sqrt{3}\right)^2}
\]

Now, I will calculate this magnitude.The intensity (magnitude) of the resulting force at point \( A \) is approximately **217.93 N**.

Would you like more details or have any questions about the calculations?

Here are some related questions you might find interesting:
1. How do you find the direction cosines of the resultant force?
2. What is the significance of using unit vectors in force analysis?
3. How would the resultant force change if \( F_{AB} \) or \( F_{AC} \) were in different directions?
4. What role does vector addition play in determining the resultant force?
5. How can the concepts used here be applied to other engineering problems, like torque or equilibrium analysis?

**Tip:** Always double-check your vector calculations by ensuring that the unit vectors are correctly normalized before applying force magnitudes.

Discover how to calculate the intensity (magnitude) of the resulting force at point A using vector analysis. This detailed solution involves determining force components, unit vectors, and vector addition principles. Learn step-by-step with practical examples.

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