Given three vectors `U=2hat(i)+3hat(j)-6hat(k), V=6hat(i)+2hat(j)+2hat(k) and W=3hat(i)-6hat(j)-2hat(k)` which of the following hold good for the vectors U, V and W ?

Options

A. U, V and W are linearly dependent
B. `(UtimesV)timesW=0`
C. U, V and W form a triplet of mutually perpendicular vectors
D. `Utimes(VtimesW)=0`

Answer

To solve the problem, we need to analyze the three vectors \( U \), \( V \), and \( W \) given as:

\[ U = 2\hat{i} + 3\hat{j} - 6\hat{k} \]
\[ V = 6\hat{i} + 2\hat{j} + 2\hat{k} \]
\[ W = 3\hat{i} - 6\hat{j} - 2\hat{k} \]

We will check the following statements:
1. \( U, V, W \) are linearly independent.
2. \( U \times V \times W = 0 \).
3. \( U, V, W \) are mutually perpendicular.
4. \( U \times V \times W = 0 \).

### Step 1: Check if \( U, V, W \) are linearly independent

To check if the vectors are linearly independent, we can calculate the scalar triple product \( U \cdot (V \times W) \). If the result is non-zero, the vectors are linearly independent.

**Calculate \( V \times W \)**:

\[ V \times W = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 2 & 2 \\ 3 & -6 & -2 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \left( 2 \cdot (-2) - 2 \cdot (-6) \right) - \hat{j} \left( 6 \cdot (-2) - 2 \cdot 3 \right) + \hat{k} \left( 6 \cdot (-6) - 2 \cdot 3 \right) \]
\[ = \hat{i} (-4 + 12) - \hat{j} (-12 - 6) + \hat{k} (-36 - 6) \]
\[ = 8\hat{i} + 18\hat{j} - 42\hat{k} \]

**Now calculate \( U \cdot (V \times W) \)**:

\[ U \cdot (V \times W) = (2\hat{i} + 3\hat{j} - 6\hat{k}) \cdot (8\hat{i} + 18\hat{j} - 42\hat{k}) \]
\[ = 2 \cdot 8 + 3 \cdot 18 - 6 \cdot 42 \]
\[ = 16 + 54 - 252 \]
\[ = 70 - 252 = -182 \]

Since \( U \cdot (V \times W) \neq 0 \), the vectors are linearly independent.

### Step 2: Check if \( U \times V \times W = 0 \)

We already calculated \( V \times W \). Now we need to calculate \( U \times (V \times W) \).

Using the result from \( V \times W \):

\[ U \times (V \times W) = U \times (8\hat{i} + 18\hat{j} - 42\hat{k}) \]

Calculating the determinant:

\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -6 \\ 8 & 18 & -42 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} (3 \cdot (-42) - (-6) \cdot 18) - \hat{j} (2 \cdot (-42) - (-6) \cdot 8) + \hat{k} (2 \cdot 18 - 3 \cdot 8) \]
\[ = \hat{i} (-126 + 108) - \hat{j} (-84 + 48) + \hat{k} (36 - 24) \]
\[ = -18\hat{i} + 36\hat{j} + 12\hat{k} \]

Since \( U \times (V \times W) \neq 0 \), this statement is false.

### Step 3: Check if \( U, V, W \) are mutually perpendicular

For the vectors to be mutually perpendicular, the dot products \( U \cdot V \), \( V \cdot W \), and \( U \cdot W \) must all be zero.

**Calculate \( U \cdot V \)**:

\[ U \cdot V = (2)(6) + (3)(2) + (-6)(2) = 12 + 6 - 12 = 6 \quad (\text{not } 0) \]

Since \( U \cdot V \neq 0 \), they are not mutually perpendicular.

### Step 4: Check if \( U \times V \times W = 0 \)

This statement is the same as the second statement we checked, and we found that it is false.

### Summary of Results:
- **Statement 1**: True (Vectors are linearly independent)
- **Statement 2**: False
- **Statement 3**: False (Not mutually perpendicular)
- **Statement 4**: False

### Final Conclusion:
The only statement that holds true is that \( U, V, W \) are linearly independent.