If `veca,vecb and vecc` are non coplanar and unit vectors such that `vecaxx(vecbxxvecc)=(vecb+vecc)/sqrt2)` then the angle between `vea and vecb` is (A) `(3pi)/4` (B) `pi/4` (C) `pi/2` (D) `pi`

Options

A. `3 pi//4`
B. `pi//4`
C. `pi//2`
D. `pi`

Explanation

To find the angle between the vectors \(\vec{a}\) and \(\vec{b}\), we will follow these steps:

### Step 1: Understand the given information
We are given that \(\vec{a}, \vec{b}, \text{ and } \vec{c}\) are non-coplanar unit vectors, and the equation is:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}} \]

### Step 2: Use the vector triple product identity
Using the vector triple product identity, we can rewrite \(\vec{a} \times (\vec{b} \times \vec{c})\) as:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} \]
Thus, we can equate:
\[ (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{\vec{b} + \vec{c}}{\sqrt{2}} \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ (\vec{a} \cdot \vec{c})\vec{b} - \frac{\vec{b}}{\sqrt{2}} = (\vec{a} \cdot \vec{b})\vec{c} + \frac{\vec{c}}{\sqrt{2}} \]
This implies:
\[ \left(\vec{a} \cdot \vec{c} - \frac{1}{\sqrt{2}}\right)\vec{b} = \left(\vec{a} \cdot \vec{b} + \frac{1}{\sqrt{2}}\right)\vec{c} \]

### Step 4: Since \(\vec{b}\) and \(\vec{c}\) are non-coplanar
For the equation to hold, the coefficients of \(\vec{b}\) and \(\vec{c}\) must be equal to zero because \(\vec{b}\) and \(\vec{c}\) are linearly independent:
1. \(\vec{a} \cdot \vec{c} - \frac{1}{\sqrt{2}} = 0\)
2. \(\vec{a} \cdot \vec{b} + \frac{1}{\sqrt{2}} = 0\)

### Step 5: Solve for dot products
From the first equation:
\[ \vec{a} \cdot \vec{c} = \frac{1}{\sqrt{2}} \]
From the second equation:
\[ \vec{a} \cdot \vec{b} = -\frac{1}{\sqrt{2}} \]

### Step 6: Find the angle between \(\vec{a}\) and \(\vec{b}\)
Using the definition of the dot product:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Since both \(\vec{a}\) and \(\vec{b}\) are unit vectors, we have:
\[ -\frac{1}{\sqrt{2}} = 1 \cdot 1 \cdot \cos \theta \]
Thus:
\[ \cos \theta = -\frac{1}{\sqrt{2}} \]

### Step 7: Find the angle \(\theta\)
The angle \(\theta\) that satisfies \(\cos \theta = -\frac{1}{\sqrt{2}}\) is:
\[ \theta = \frac{3\pi}{4} \]

### Conclusion
Therefore, the angle between \(\vec{a}\) and \(\vec{b}\) is:
\[ \boxed{\frac{3\pi}{4}} \]