If ` veca and vecb` are unit vectors inclined to x-axis at angle ` 30^(@) and 120^(@)` then ` |veca +vecb|` equals

Options

A. `sqrt(2//3)`
B. `sqrt2`
C. `sqrt3`
D. 2

Explanation

To find the magnitude of the sum of two unit vectors \( \vec{a} \) and \( \vec{b} \) that are inclined to the x-axis at angles \( 30^\circ \) and \( 120^\circ \) respectively, we can follow these steps:

### Step 1: Represent the vectors in component form
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we can express them in terms of their components:
- For \( \vec{a} \):
\[ \vec{a} = \cos(30^\circ) \hat{i} + \sin(30^\circ) \hat{j} = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \]
- For \( \vec{b} \):
\[ \vec{b} = \cos(120^\circ) \hat{i} + \sin(120^\circ) \hat{j} = -\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \]

### Step 2: Add the vectors
Now, we can add the two vectors:
\[ \vec{a} + \vec{b} = \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) \hat{i} + \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) \hat{j} \]
Calculating the components:
- For the \( \hat{i} \) component:
\[ \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2} \]
- For the \( \hat{j} \) component:
\[ \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \]
Thus,
\[ \vec{a} + \vec{b} = \left( \frac{\sqrt{3} - 1}{2} \right) \hat{i} + \left( \frac{1 + \sqrt{3}}{2} \right) \hat{j} \]

### Step 3: Calculate the magnitude of the resultant vector
The magnitude of \( \vec{a} + \vec{b} \) is given by:
\[ |\vec{a} + \vec{b}| = \sqrt{\left( \frac{\sqrt{3} - 1}{2} \right)^2 + \left( \frac{1 + \sqrt{3}}{2} \right)^2} \]

Calculating each term:
1. For the \( \hat{i} \) component:
\[ \left( \frac{\sqrt{3} - 1}{2} \right)^2 = \frac{(\sqrt{3} - 1)^2}{4} = \frac{3 - 2\sqrt{3} + 1}{4} = \frac{4 - 2\sqrt{3}}{4} = 1 - \frac{\sqrt{3}}{2} \]
2. For the \( \hat{j} \) component:
\[ \left( \frac{1 + \sqrt{3}}{2} \right)^2 = \frac{(1 + \sqrt{3})^2}{4} = \frac{1 + 2\sqrt{3} + 3}{4} = \frac{4 + 2\sqrt{3}}{4} = 1 + \frac{\sqrt{3}}{2} \]

Now, adding these two results:
\[ |\vec{a} + \vec{b}|^2 = \left( 1 - \frac{\sqrt{3}}{2} \right) + \left( 1 + \frac{\sqrt{3}}{2} \right) = 2 \]

### Step 4: Final calculation of the magnitude
Taking the square root gives:
\[ |\vec{a} + \vec{b}| = \sqrt{2} \]

### Final Answer
Thus, the magnitude \( |\vec{a} + \vec{b}| \) equals \( \sqrt{2} \).

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