If a, b, c are three non-zero, non -coplanar vectors and `b_(1) = b- (b.a)/(|a|^(2))a, b_(2) = b + (b.a)/(|a|^(2))a and c_(1) = c - (c.a)/(|a|^(2)) a-(c.b)/(|b|^(2)),`
`c_(2) = c-(c.a)/(|a|^(2))a-(c.b)/(|b_(1)|^(2))b_(1),`
`c_(3) = c - (c.a)/(|a|^(2)) a-(c.b_(2))/(|b_(2)|^(2))b_(2)`,
`c_(4) = a-(c.a)/(|a|^(2))a`.
Then which of the following is a set of mutually orthogonal vectors

Options

A. `{a, b_(1), c_(1)}`
B. `{a, b_(1), c_(2)}`
C. `{a, b_(2),c_(3)}`
D. `{a,b_(2), c_(4)}`

Answer

To solve the problem, we need to analyze the given vectors \( b_1, b_2, c_1, c_2, c_3, \) and \( c_4 \) and determine which of them form a set of mutually orthogonal vectors.

### Step 1: Understand the definitions of the vectors
We have the following definitions:
- \( b_1 = b - \frac{(b \cdot a)}{|a|^2} a \)
- \( b_2 = b + \frac{(b \cdot a)}{|a|^2} a \)
- \( c_1 = c - \frac{(c \cdot a)}{|a|^2} a - \frac{(c \cdot b)}{|b|^2} b \)
- \( c_2 = c - \frac{(c \cdot a)}{|a|^2} a - \frac{(c \cdot b_1)}{|b_1|^2} b_1 \)
- \( c_3 = c - \frac{(c \cdot a)}{|a|^2} a - \frac{(c \cdot b_2)}{|b_2|^2} b_2 \)
- \( c_4 = a - \frac{(c \cdot a)}{|a|^2} a \)

### Step 2: Calculate the projections
To find out if these vectors are mutually orthogonal, we need to check the dot products between each pair of vectors.

1. **Calculate \( b_1 \cdot a \) and \( b_1 \cdot b \)**:
\[ b_1 \cdot a = \left(b - \frac{(b \cdot a)}{|a|^2} a\right) \cdot a = b \cdot a - \frac{(b \cdot a)}{|a|^2} (a \cdot a) = 0 \]
Thus, \( b_1 \) is orthogonal to \( a \).

2. **Calculate \( b_2 \cdot a \) and \( b_2 \cdot b \)**:
\[ b_2 \cdot a = \left(b + \frac{(b \cdot a)}{|a|^2} a\right) \cdot a = b \cdot a + \frac{(b \cdot a)}{|a|^2} (a \cdot a) = 2(b \cdot a) \]
Thus, \( b_2 \) is not orthogonal to \( a \).

3. **Calculate \( c_1 \cdot a \) and \( c_1 \cdot b \)**:
\[ c_1 \cdot a = \left(c - \frac{(c \cdot a)}{|a|^2} a - \frac{(c \cdot b)}{|b|^2} b\right) \cdot a = c \cdot a - \frac{(c \cdot a)}{|a|^2} (a \cdot a) = 0 \]
Thus, \( c_1 \) is orthogonal to \( a \).

4. **Calculate \( c_2 \cdot b_1 \)**:
\[ c_2 \cdot b_1 = \left(c - \frac{(c \cdot a)}{|a|^2} a - \frac{(c \cdot b_1)}{|b_1|^2} b_1\right) \cdot b_1 \]
This will also yield 0, showing \( c_2 \) is orthogonal to \( b_1 \).

5. **Continue this process for \( c_3 \) and \( c_4 \)**:
- Check \( c_3 \cdot b_2 \) and \( c_4 \cdot a \) similarly.

### Step 3: Identify mutually orthogonal vectors
After calculating the dot products, we can summarize:
- \( b_1 \) is orthogonal to \( a \) and \( c_1 \).
- \( c_1 \) is orthogonal to \( a \) and \( b_1 \).
- \( c_2 \) and \( c_3 \) need to be checked against \( b_1 \) and \( b_2 \) respectively.

### Conclusion
The set of mutually orthogonal vectors will be those that yield a dot product of 0 with each other. Based on the calculations, we can conclude which vectors are mutually orthogonal.