Let `hata` be a unit vector and `hatb` a non zero vector non parallel to `veca`. Find the angles of the triangle tow sides of which are represented by the vectors. `sqrt(3)(hatxxvecb)and vecb-(hata.vecb)hata`

Options

A. `tan ^(-1)(sqrt3)`
B. `tan ^(-1)(1//sqrt3)`
C. `cot^(-1)(0)`
D. tant^(-1)(1)`

Explanation

To solve the problem, we need to find the angles of a triangle whose two sides are represented by the vectors \( \sqrt{3} \hat{b} \) and \( \hat{b} - (\hat{a} \cdot \hat{b}) \hat{a} \).

### Step 1: Identify the vectors
Let:
- \( \vec{v_1} = \sqrt{3} \hat{b} \)
- \( \vec{v_2} = \hat{b} - (\hat{a} \cdot \hat{b}) \hat{a} \)

### Step 2: Check for orthogonality
To find the angle between the two vectors, we first check if they are orthogonal. If \( \vec{v_1} \cdot \vec{v_2} = 0 \), then the angle between them is \( 90^\circ \).

Calculating the dot product:
\[ \vec{v_1} \cdot \vec{v_2} = \sqrt{3} \hat{b} \cdot \left(\hat{b} - (\hat{a} \cdot \hat{b}) \hat{a}\right) \]
\[ = \sqrt{3} (\hat{b} \cdot \hat{b}) - \sqrt{3} (\hat{b} \cdot \hat{a})(\hat{a} \cdot \hat{b}) \]
Since \( \hat{b} \cdot \hat{b} = 1 \) (because \( \hat{b} \) is a unit vector), we have:
\[ = \sqrt{3} - \sqrt{3} (\hat{b} \cdot \hat{a})^2 \]
Setting this equal to zero gives:
\[ \sqrt{3} - \sqrt{3} (\hat{b} \cdot \hat{a})^2 = 0 \]
\[ \Rightarrow 1 - (\hat{b} \cdot \hat{a})^2 = 0 \]
\[ \Rightarrow (\hat{b} \cdot \hat{a})^2 = 1 \]
This implies \( \hat{b} \cdot \hat{a} = 1 \) or \( \hat{b} \cdot \hat{a} = -1 \). Since \( \hat{b} \) is not parallel to \( \hat{a} \), we can conclude that they are orthogonal.

### Step 3: Use the sine law
Using the sine law in triangle, we can express the relationship between the sides and angles:
\[ \frac{\vec{v_2}}{\sin(\theta)} = \frac{\vec{v_1}}{\sin(90^\circ)} \]
This simplifies to:
\[ \frac{\hat{b} - (\hat{a} \cdot \hat{b}) \hat{a}}{\sin(\theta)} = \frac{\sqrt{3} \hat{b}}{1} \]
From this, we can derive:
\[ \sin(\theta) = \frac{\hat{b} - (\hat{a} \cdot \hat{b}) \hat{a}}{\sqrt{3} \hat{b}} \]

### Step 4: Calculate the angle
Using the relationship \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can find \( \theta \):
\[ \tan(\theta) = \frac{1}{\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6} \]

### Conclusion
The angles of the triangle formed by the two sides represented by the vectors \( \sqrt{3} \hat{b} \) and \( \hat{b} - (\hat{a} \cdot \hat{b}) \hat{a} \) is \( \frac{\pi}{6} \) radians or \( 30^\circ \).