If a, b, c are non-zero, non-collinear vectors such that a vectors such that a vector `p=abcos(2pi-(a,c))c and aq=ac cos(pi-(a, c))` then b+q is

Options

A. (a)parallel to a
B. (b)perpendicular to a
C. (c)coplanar with b and c
D. (d)coplanar with a and c

Answer

To solve the problem, we need to analyze the vectors \( p \) and \( q \) given in the question and find \( b + q \).

### Step-by-Step Solution:

1. **Understanding the Vectors:**
We have two vectors defined as:
\[ p = ab \cos(2\pi - (a, c)) c \]
\[ q = ac \cos(\pi - (a, c)) \]

2. **Using the Cosine Identity:**
We know that:
\[ \cos(2\pi - x) = \cos(x) \quad \text{and} \quad \cos(\pi - x) = -\cos(x) \]
Therefore, we can rewrite \( p \) and \( q \):
\[ p = ab \cos((a, c)) c \]
\[ q = -ac \cos((a, c) \]

3. **Finding \( b + q \):**
Now we need to find \( b + q \):
\[ b + q = b - ac \cos((a, c) \]

4. **Dot Product with Vector \( a \):**
To analyze \( p + q \), we can take the dot product with vector \( a \):
\[ a \cdot p = a \cdot (ab \cos((a, c)) c) = a \cdot a b \cos((a, c)) (a \cdot c) \]
\[ a \cdot q = a \cdot (-ac \cos((a, c)) = -a \cdot a c \cos((a, c)) \]

5. **Combining the Dot Products:**
Now we combine these:
\[ a \cdot (p + q) = a \cdot p + a \cdot q = (a \cdot a) b \cos((a, c)) (a \cdot c) - (a \cdot a) c \cos((a, c)) \]

6. **Simplifying the Expression:**
Since \( a \cdot a \) is a scalar and can be factored out:
\[ a \cdot (p + q) = (a \cdot a) \cos((a, c)) (b \cdot c - c) \]

7. **Conclusion:**
Since \( a \cdot (p + q) = 0 \), we conclude that \( p + q \) is perpendicular to vector \( a \). Therefore, \( b + q \) must also be perpendicular to vector \( a \).

### Final Result:
Thus, the answer is that \( b + q \) is perpendicular to vector \( a \).