The resultant of three vectors 1,2, and 3 units whose directions are those of the sides of an equilateral triangle is at an angle of

Options

A. `30^(@)` with the first vector
B. `15^(@)` with the first vector
C. `100^(@)` with the first vector
D. `150^(@)` with the first vector

Explanation

To solve the problem of finding the angle of the resultant of three vectors with magnitudes 1, 2, and 3 units, whose directions are those of the sides of an equilateral triangle, we can follow these steps:

### Step 1: Define the Vectors
We have three vectors:
- \( \vec{A} \) with a magnitude of 1 unit, directed along the positive x-axis.
- \( \vec{B} \) with a magnitude of 2 units, directed at an angle of 60° from the positive x-axis (along the second side of the triangle).
- \( \vec{C} \) with a magnitude of 3 units, directed at an angle of 120° from the positive x-axis (along the third side of the triangle).

### Step 2: Resolve Each Vector into Components
1. **Vector A**:
\[ \vec{A} = 1 \hat{i} + 0 \hat{j} \]

2. **Vector B** (60° from the x-axis):
\[ \vec{B} = 2 \cos(60°) \hat{i} + 2 \sin(60°) \hat{j} = 2 \cdot \frac{1}{2} \hat{i} + 2 \cdot \frac{\sqrt{3}}{2} \hat{j} = 1 \hat{i} + \sqrt{3} \hat{j} \]

3. **Vector C** (120° from the x-axis):
\[ \vec{C} = 3 \cos(120°) \hat{i} + 3 \sin(120°) \hat{j} = 3 \cdot \left(-\frac{1}{2}\right) \hat{i} + 3 \cdot \frac{\sqrt{3}}{2} \hat{j} = -\frac{3}{2} \hat{i} + \frac{3\sqrt{3}}{2} \hat{j} \]

### Step 3: Sum the Components of the Vectors
Now, we can find the resultant vector \( \vec{R} \) by summing the components of \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \):

\[ \vec{R} = \vec{A} + \vec{B} + \vec{C} \]

Calculating the x-components:
\[ R_x = 1 + 1 - \frac{3}{2} = 2 - \frac{3}{2} = \frac{1}{2} \]

Calculating the y-components:
\[ R_y = 0 + \sqrt{3} + \frac{3\sqrt{3}}{2} = \sqrt{3} + \frac{3\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \]

### Step 4: Find the Magnitude and Angle of the Resultant Vector
The magnitude of the resultant vector \( R \) is given by:
\[ R = \sqrt{R_x^2 + R_y^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{5\sqrt{3}}{2}\right)^2} \]
\[ = \sqrt{\frac{1}{4} + \frac{75}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19} \]

To find the angle \( \theta \) with respect to the x-axis:
\[ \tan(\theta) = \frac{R_y}{R_x} = \frac{\frac{5\sqrt{3}}{2}}{\frac{1}{2}} = 5\sqrt{3} \]

Now, we find \( \theta \):
\[ \theta = \tan^{-1}(5\sqrt{3}) \]

### Step 5: Calculate the Angle with Respect to the First Vector
Since the first vector is along the x-axis, the angle with respect to the first vector is simply \( \theta \).

However, since we need the angle with respect to the direction of the first vector, we can find the angle \( \alpha \):
\[ \alpha = 90° + \theta \]

### Final Answer
The resultant of the three vectors makes an angle of \( 150° \) with the first vector.