If `(a/costheta)=(b/(cos(theta+(2pi/3)))=(c/cos(theta+(4pi/3)))` then find angle between vectors `ahati+bhatj+chatk` and `bhati+chatj+ahatk` if `(theta= (2pi/9))` and `a^2+b^2+c^2=1`, is

Options

A. `(pi/3)`
B. `(pi/6)`
C. `(2pi/3)`
D. `(5pi/6)`

Answer

To solve the problem, we need to find the angle between the vectors \( \mathbf{A} = a \hat{i} + b \hat{j} + c \hat{k} \) and \( \mathbf{B} = b \hat{i} + c \hat{j} + a \hat{k} \) given the conditions provided.

### Step 1: Set up the equations

Given:
\[ \frac{a}{\cos \theta} = \frac{b}{\cos(\theta + \frac{2\pi}{3})} = \frac{c}{\cos(\theta + \frac{4\pi}{3})} = k \]

From this, we can express \( a, b, c \) in terms of \( k \):
\[ a = k \cos \theta \]
\[ b = k \cos\left(\theta + \frac{2\pi}{3}\right) \]
\[ c = k \cos\left(\theta + \frac{4\pi}{3}\right) \]

### Step 2: Substitute \( \theta \)

Given \( \theta = \frac{2\pi}{9} \), we substitute this into the equations for \( a, b, c \):
\[ a = k \cos\left(\frac{2\pi}{9}\right) \]
\[ b = k \cos\left(\frac{2\pi}{9} + \frac{2\pi}{3}\right) = k \cos\left(\frac{2\pi}{9} + \frac{6\pi}{9}\right) = k \cos\left(\frac{8\pi}{9}\right) \]
\[ c = k \cos\left(\frac{2\pi}{9} + \frac{4\pi}{3}\right) = k \cos\left(\frac{2\pi}{9} + \frac{12\pi}{9}\right) = k \cos\left(\frac{14\pi}{9}\right) \]

### Step 3: Use the identity for cosine

Using the identity \( \cos(x + y) = \cos x \cos y - \sin x \sin y \), we can find \( b \) and \( c \):
- \( \cos\left(\frac{8\pi}{9}\right) = -\cos\left(\frac{\pi}{9}\right) \)
- \( \cos\left(\frac{14\pi}{9}\right) = -\cos\left(\frac{4\pi}{9}\right) \)

Thus, we have:
\[ b = -k \cos\left(\frac{\pi}{9}\right) \]
\[ c = -k \cos\left(\frac{4\pi}{9}\right) \]

### Step 4: Use the condition \( a^2 + b^2 + c^2 = 1 \)

Now we substitute \( a, b, c \) into the equation:
\[ (k \cos\left(\frac{2\pi}{9}\right))^2 + (-k \cos\left(\frac{\pi}{9}\right))^2 + (-k \cos\left(\frac{4\pi}{9}\right))^2 = 1 \]
\[ k^2 \left( \cos^2\left(\frac{2\pi}{9}\right) + \cos^2\left(\frac{\pi}{9}\right) + \cos^2\left(\frac{4\pi}{9}\right) \right) = 1 \]

Let \( S = \cos^2\left(\frac{2\pi}{9}\right) + \cos^2\left(\frac{\pi}{9}\right) + \cos^2\left(\frac{4\pi}{9}\right) \). Then:
\[ k^2 S = 1 \implies k = \frac{1}{\sqrt{S}} \]

### Step 5: Calculate the dot product \( \mathbf{A} \cdot \mathbf{B} \)

Now we can find the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = a b + b c + c a \]
Substituting the values:
\[ = k^2 \left( \cos\left(\frac{2\pi}{9}\right)(-\cos\left(\frac{\pi}{9}\right)) + (-k \cos\left(\frac{\pi}{9}\right))(-k \cos\left(\frac{4\pi}{9}\right)) + (-k \cos\left(\frac{4\pi}{9}\right))(k \cos\left(\frac{2\pi}{9}\right)) \right) \]

### Step 6: Find the angle between the vectors

Using the formula for the angle between two vectors:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

Since \( |\mathbf{A}| = |\mathbf{B}| = 1 \) (from the condition \( a^2 + b^2 + c^2 = 1 \)), we can find the angle \( \theta \).

### Final Step: Conclusion

After calculating the dot product and substituting back into the cosine formula, we find:
\[ \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3} \]

Thus, the angle between the vectors \( \mathbf{A} \) and \( \mathbf{B} \) is \( \frac{2\pi}{3} \).