If `veca and vecb` are unit vectors such that `(veca +vecb). (2veca + 3vecb)xx(3veca - 2vecb)=vec0` then angle between `veca and vecb` is

Options

A. 0
B. `pi//2`
C. `pi`
D. indeterminate

Explanation

To solve the problem step by step, we start with the given equation:

\[ (\vec{a} + \vec{b}) \cdot ((2\vec{a} + 3\vec{b}) \times (3\vec{a} - 2\vec{b})) = \vec{0} \]

### Step 1: Simplify the Cross Product
We need to simplify the expression \( (2\vec{a} + 3\vec{b}) \times (3\vec{a} - 2\vec{b}) \).

Using the distributive property of the cross product:

\[ (2\vec{a} + 3\vec{b}) \times (3\vec{a} - 2\vec{b}) = 2\vec{a} \times 3\vec{a} + 2\vec{a} \times (-2\vec{b}) + 3\vec{b} \times 3\vec{a} + 3\vec{b} \times (-2\vec{b}) \]

Since the cross product of any vector with itself is zero, we have:

\[ 2\vec{a} \times 3\vec{a} = \vec{0} \quad \text{and} \quad 3\vec{b} \times (-2\vec{b}) = \vec{0} \]

Thus, we are left with:

\[ = -4\vec{a} \times \vec{b} + 9\vec{b} \times \vec{a} \]

Using the property \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \):

\[ = -4\vec{a} \times \vec{b} - 9\vec{a} \times \vec{b} = -13\vec{a} \times \vec{b} \]

### Step 2: Substitute Back into the Original Equation
Now substituting back into the original equation, we have:

\[ (\vec{a} + \vec{b}) \cdot (-13\vec{a} \times \vec{b}) = 0 \]

### Step 3: Factor Out the Scalar
This can be simplified to:

\[ -13(\vec{a} + \vec{b}) \cdot (\vec{a} \times \vec{b}) = 0 \]

Since \(-13\) is a non-zero scalar, we can divide both sides by \(-13\):

\[ (\vec{a} + \vec{b}) \cdot (\vec{a} \times \vec{b}) = 0 \]

### Step 4: Analyze the Result
The expression \( (\vec{a} + \vec{b}) \cdot (\vec{a} \times \vec{b}) = 0 \) implies that the vector \( \vec{a} \times \vec{b} \) is perpendicular to \( \vec{a} + \vec{b} \).

### Step 5: Conclusion about the Angle
Since \( \vec{a} \times \vec{b} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \), the angle between \( \vec{a} \) and \( \vec{b} \) can be any value that satisfies the condition of being unit vectors. Therefore, the angle between \( \vec{a} \) and \( \vec{b} \) is indeterminate.

### Final Answer
The angle between \( \vec{a} \) and \( \vec{b} \) is indeterminate.

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