Let a, b, c are non-zero unit vectors inclined pairwise with the same angle `theta`, p, q, r are non zero scalars satisfying `atimesb+btimesc=pa+qb+rc` Q. The value of `((q)/(p)+2costheta)` is

Options

A. `1`
B. `0`
C. `2[a b c]`
D. None of these

Answer

To solve the problem, we start with the given vectors and scalars. We have non-zero unit vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) that are pairwise inclined at the same angle \( \theta \), and scalars \( p, q, r \) such that:

\[ \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} = p\mathbf{a} + q\mathbf{b} + r\mathbf{c} \]

We need to find the value of \( \frac{q}{p} + 2\cos\theta \).

### Step 1: Dot Product with \( \mathbf{a} \)

Taking the dot product of both sides with \( \mathbf{a} \):

\[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} + (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = p(\mathbf{a} \cdot \mathbf{a}) + q(\mathbf{b} \cdot \mathbf{a}) + r(\mathbf{c} \cdot \mathbf{a}) \]

Since \( \mathbf{a} \times \mathbf{b} \) is orthogonal to \( \mathbf{a} \), the first term is zero:

\[ 0 + (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = p + q\cos\theta + r\cos\theta \]

Thus, we have:

\[ (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = p + (q + r)\cos\theta \tag{1} \]

### Step 2: Dot Product with \( \mathbf{b} \)

Now, taking the dot product with \( \mathbf{b} \):

\[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} + (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{b} = p(\mathbf{b} \cdot \mathbf{a}) + q(\mathbf{b} \cdot \mathbf{b}) + r(\mathbf{b} \cdot \mathbf{c}) \]

The first term is zero, and the second term is also zero because \( \mathbf{b} \times \mathbf{b} = 0 \):

\[ 0 + 0 = p\cos\theta + q + r\cos\theta \]

Thus, we have:

\[ p\cos\theta + q + r\cos\theta = 0 \tag{2} \]

### Step 3: Dot Product with \( \mathbf{c} \)

Now, taking the dot product with \( \mathbf{c} \):

\[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} + (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{c} = p(\mathbf{c} \cdot \mathbf{a}) + q(\mathbf{c} \cdot \mathbf{b}) + r(\mathbf{c} \cdot \mathbf{c}) \]

Again, the first term is zero, and the second term is zero:

\[ 0 + 0 = p\cos\theta + q\cos\theta + r \]

Thus, we have:

\[ p\cos\theta + q\cos\theta + r = 0 \tag{3} \]

### Step 4: Equating Equations

From equations (1), (2), and (3), we can equate the right-hand sides. From (2) and (3):

Setting the right-hand sides of (2) and (3) equal gives:

\[ p + (q + r)\cos\theta = p\cos\theta + q + r\cos\theta \]

This simplifies to:

\[ p = r \tag{4} \]

### Step 5: Substitute \( r \) into Equation (2)

Substituting \( r = p \) into equation (2):

\[ p\cos\theta + q + p\cos\theta = 0 \]

This simplifies to:

\[ 2p\cos\theta + q = 0 \]

Thus, we have:

\[ q = -2p\cos\theta \tag{5} \]

### Step 6: Calculate \( \frac{q}{p} + 2\cos\theta \)

Now we can find \( \frac{q}{p} + 2\cos\theta \):

\[ \frac{q}{p} = \frac{-2p\cos\theta}{p} = -2\cos\theta \]

Thus,

\[ \frac{q}{p} + 2\cos\theta = -2\cos\theta + 2\cos\theta = 0 \]

### Final Answer

The value of \( \frac{q}{p} + 2\cos\theta \) is:

\[ \boxed{0} \]