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

### Step 1: Understand the Geometry
We have three vectors forming an equilateral triangle. Let's denote the vectors as:
- \( \vec{A_1} \) = 1 unit
- \( \vec{A_2} \) = 2 units
- \( \vec{A_3} \) = 3 units

The angles between each vector are \( 60^\circ \).

### Step 2: Break Down the Vectors into Components
Assuming \( \vec{A_1} \) is along the positive x-axis:
- \( \vec{A_1} = 1 \hat{i} + 0 \hat{j} \)
- \( \vec{A_2} \) makes an angle of \( 60^\circ \) with the x-axis:
- \( \vec{A_2} = 2 \cos(60^\circ) \hat{i} + 2 \sin(60^\circ) \hat{j} = 2 \cdot \frac{1}{2} \hat{i} + 2 \cdot \frac{\sqrt{3}}{2} \hat{j} = 1 \hat{i} + \sqrt{3} \hat{j} \)
- \( \vec{A_3} \) makes an angle of \( 120^\circ \) with the x-axis:
- \( \vec{A_3} = 3 \cos(120^\circ) \hat{i} + 3 \sin(120^\circ) \hat{j} = 3 \cdot \left(-\frac{1}{2}\right) \hat{i} + 3 \cdot \frac{\sqrt{3}}{2} \hat{j} = -\frac{3}{2} \hat{i} + \frac{3\sqrt{3}}{2} \hat{j} \)

### Step 3: Calculate the Resultant Vector
Now, we can find the resultant vector \( \vec{R} \) by adding the components of the three vectors:
\[ \vec{R} = \vec{A_1} + \vec{A_2} + \vec{A_3} \]
Calculating the x-components:
\[ R_x = 1 + 1 - \frac{3}{2} = 2 - \frac{3}{2} = \frac{1}{2} \]
Calculating the y-components:
\[ R_y = 0 + \sqrt{3} + \frac{3\sqrt{3}}{2} = \sqrt{3} + \frac{3\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \]
Thus, the resultant vector is:
\[ \vec{R} = \frac{1}{2} \hat{i} + \frac{5\sqrt{3}}{2} \hat{j} \]

### Step 4: Find the Angle with the First Vector
To find the angle \( \theta \) between \( \vec{R} \) and \( \vec{A_1} \), we use the dot product:
\[ \cos(\theta) = \frac{\vec{A_1} \cdot \vec{R}}{|\vec{A_1}| |\vec{R}|} \]
Calculating the dot product:
\[ \vec{A_1} \cdot \vec{R} = (1)(\frac{1}{2}) + (0)(\frac{5\sqrt{3}}{2}) = \frac{1}{2} \]
Calculating the magnitudes:
\[ |\vec{A_1}| = 1 \]
\[ |\vec{R}| = \sqrt{(\frac{1}{2})^2 + (\frac{5\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{75}{4}} = \sqrt{19} \]
Now substituting into the cosine formula:
\[ \cos(\theta) = \frac{\frac{1}{2}}{1 \cdot \sqrt{19}} = \frac{1}{2\sqrt{19}} \]
Now, we find \( \theta \):
Using the inverse cosine function:
\[ \theta = \cos^{-1}\left(\frac{1}{2\sqrt{19}}\right) \]

### Step 5: Determine the Angle
To find the angle in degrees, we can calculate \( \theta \) using a calculator.

### Final Answer
After calculating, we find that the angle made by the resultant with the first vector is approximately \( 150^\circ \).

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