Given the following informations about the non-zero vectors A, B and C
`(i)(AtimesB)timesA=0 : (ii)B*B=4`
`(iii) A*B=-6 : (iv)B*C=6`
which one of the following holds good?

Options

A. `AtimesB=0`
B. `A*(BtimesC)=0`
C. `A*A=8`
D. `A*C=-1`

Answer

To solve the problem step by step, we will analyze the given conditions and derive the necessary conclusions.

### Given Information:
1. \( A \times (B \times A) = 0 \)
2. \( B \cdot B = 4 \)
3. \( A \cdot B = -6 \)
4. \( B \cdot C = 6 \)

### Step 1: Analyze the first condition
From the first condition, we have:
\[ A \times (B \times A) = 0 \]

Using the vector triple product identity:
\[ A \times (B \times A) = (A \cdot A)B - (A \cdot B)A \]

Setting this equal to zero gives:
\[ (A \cdot A)B - (A \cdot B)A = 0 \]

### Step 2: Substitute known values
Let \( A \cdot A = |A|^2 \) and \( A \cdot B = -6 \):
\[ |A|^2 B + 6A = 0 \]

Rearranging gives:
\[ |A|^2 B = -6A \]

### Step 3: Express A in terms of B
From the equation \( |A|^2 B = -6A \), we can express \( A \) as:
\[ A = -\frac{|A|^2}{6} B \]

This implies that \( A \) is parallel to \( B \).

### Step 4: Check the implications of parallel vectors
If \( A \) is parallel to \( B \), then:
\[ A \times B = 0 \]

### Step 5: Analyze the second condition
Now, we take the equation \( |A|^2 B + 6A = 0 \) and take the cross product with \( C \):
\[ |A|^2 (B \times C) + 6(A \times C) = 0 \]

### Step 6: Dot product with A
Taking the dot product with \( A \):
\[ |A|^2 (A \cdot (B \times C)) + 6(A \cdot (A \times C)) = 0 \]

Since \( A \times C \) is perpendicular to \( A \), \( A \cdot (A \times C) = 0 \):
\[ |A|^2 (A \cdot (B \times C)) = 0 \]

This implies:
\[ A \cdot (B \times C) = 0 \]

### Step 7: Conclusion from the second condition
This means that \( A \) is orthogonal to \( B \times C \).

### Step 8: Analyze the third condition
From \( A \cdot B = -6 \):
Using the formula:
\[ A \cdot B = |A||B|\cos(\theta) \]
Since \( A \) is parallel to \( B \), the angle \( \theta \) must be \( \pi \) (180 degrees):
\[ |A| \cdot |B| \cdot (-1) = -6 \]

Given \( |B|^2 = 4 \) implies \( |B| = 2 \):
\[ |A| \cdot 2 = 6 \]
\[ |A| = 3 \]
Thus, \( |A|^2 = 9 \).

### Step 9: Check consistency with the given conditions
However, the problem states that \( A \cdot A = 8 \), which contradicts our derived value of \( 9 \). Therefore, this condition is false.

### Step 10: Analyze the fourth condition
Using \( |A|^2 B + 6A = 0 \) and taking the dot product with \( C \):
\[ |A|^2 (B \cdot C) + 6(A \cdot C) = 0 \]
Substituting \( |A|^2 = 9 \) and \( B \cdot C = 6 \):
\[ 9 \cdot 6 + 6(A \cdot C) = 0 \]
\[ 54 + 6(A \cdot C) = 0 \]
\[ 6(A \cdot C) = -54 \]
\[ A \cdot C = -9 \]

### Step 11: Conclusion from the fourth condition
The problem states \( A \cdot C = -1 \), which is also false.

### Final Summary
The only conditions that hold true from the analysis are:
1. \( A \) is parallel to \( B \).
2. \( A \cdot (B \times C) = 0 \).