Let a, b and c be non-zero vectors and `|a|=1` and `r` is a non-zero vector such that `r times a=b` and `r*a=1`, then

Options

A. `abotb`
B. `rbotb`
C. `r*a=(1-[a b c])/(a*b)`
D. [r a b]=0

Answer

To solve the problem step by step, we will analyze the given information and derive the required conclusions.

### Given:
1. \( |a| = 1 \) (vector \( a \) is a unit vector)
2. \( r \) is a non-zero vector such that:
- \( r \times a = b \)
- \( r \cdot a = 1 \)

### To Prove:
1. Whether \( a \) is perpendicular to \( b \)
2. Whether \( r \) is perpendicular to \( b \)
3. The value of \( r \cdot a \)
4. The scalar triple product \( r \cdot (a \times b) \)

### Step 1: Verify if \( a \) is perpendicular to \( b \)

From the equation \( r \times a = b \), we can take the dot product of both sides with \( a \):

\[ (r \times a) \cdot a = b \cdot a \]

The left-hand side simplifies to zero because the cross product \( r \times a \) is perpendicular to both \( r \) and \( a \):

\[ 0 = b \cdot a \]

Thus, \( a \) is perpendicular to \( b \).

### Step 2: Verify if \( r \) is perpendicular to \( b \)

We can use the fact that \( r \times a = b \). Taking the cross product of both sides with \( b \):

\[ r \times (a \times b) = b \times b \]

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

\[ r \times (a \times b) = 0 \]

Using the vector triple product identity \( x \times (y \times z) = (x \cdot z)y - (x \cdot y)z \):

\[ r \times (a \times b) = (r \cdot b)a - (r \cdot a)b \]

Setting this equal to zero gives:

\[ (r \cdot b)a - (r \cdot a)b = 0 \]

Since \( r \cdot a = 1 \):

\[ (r \cdot b)a - b = 0 \]

This implies:

\[ (r \cdot b)a = b \]

Since \( a \) is a unit vector and \( b \) is non-zero, we conclude that \( r \cdot b = 0 \). Therefore, \( r \) is perpendicular to \( b \).

### Step 3: Value of \( r \cdot a \)

From the given information, we have:

\[ r \cdot a = 1 \]

### Step 4: Scalar triple product \( r \cdot (a \times b) \)

The scalar triple product can be expressed as:

\[ r \cdot (a \times b) = b \cdot (r \times a) \]

Since \( r \times a = b \):

\[ r \cdot (a \times b) = b \cdot b = |b|^2 \]

Since \( b \) is a non-zero vector, \( |b|^2 \) is not equal to zero.

### Conclusion:
1. \( a \) is perpendicular to \( b \) (True).
2. \( r \) is perpendicular to \( b \) (True).
3. \( r \cdot a = 1 \) (True).
4. The scalar triple product \( r \cdot (a \times b) = |b|^2 \) (True).