If `a = (2, 3, 5), b = (3, -6,2), c = (6, 2,-3)` then `a xx b =...?and b xxc = ..?` and `(a xx b) xx c = a xx (b xx c) =0` . True or False

Answer

To solve the problem, we need to calculate the cross products of the vectors \( a \), \( b \), and \( c \) as given:

1. **Given Vectors:**
- \( a = (2, 3, 5) \)
- \( b = (3, -6, 2) \)
- \( c = (6, 2, -3) \)

2. **Calculate \( a \times b \):**
The cross product \( a \times b \) can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of vectors \( a \) and \( b \).

\[ a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ 3 & -6 & 2 \end{vmatrix} \]

Expanding this determinant:

\[ a \times b = \hat{i} \begin{vmatrix} 3 & 5 \\ -6 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 5 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 3 & -6 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

- For \( \hat{i} \):
\[ 3 \cdot 2 - (-6) \cdot 5 = 6 + 30 = 36 \]

- For \( \hat{j} \):
\[ 2 \cdot 2 - 3 \cdot 5 = 4 - 15 = -11 \quad \Rightarrow \quad -(-11) = 11 \]

- For \( \hat{k} \):
\[ 2 \cdot (-6) - 3 \cdot 3 = -12 - 9 = -21 \]

Therefore, combining these results:
\[ a \times b = 36 \hat{i} + 11 \hat{j} - 21 \hat{k} = (36, 11, -21) \]

3. **Calculate \( b \times c \):**
Similarly, we calculate \( b \times c \):

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

Expanding this determinant:

\[ b \times c = \hat{i} \begin{vmatrix} -6 & 2 \\ 2 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 \\ 6 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -6 \\ 6 & 2 \end{vmatrix} \]

Calculating these 2x2 determinants:

- For \( \hat{i} \):
\[ (-6)(-3) - (2)(2) = 18 - 4 = 14 \]

- For \( \hat{j} \):
\[ 3 \cdot (-3) - 2 \cdot 6 = -9 - 12 = -21 \quad \Rightarrow \quad -(-21) = 21 \]

- For \( \hat{k} \):
\[ 3 \cdot 2 - (-6) \cdot 6 = 6 + 36 = 42 \]

Therefore, combining these results:
\[ b \times c = 14 \hat{i} + 21 \hat{j} + 42 \hat{k} = (14, 21, 42) \]

4. **Check if \( (a \times b) \times c = a \times (b \times c) = 0 \):**
We need to calculate \( (a \times b) \times c \) and \( a \times (b \times c) \).

First, we calculate \( (a \times b) \times c \):
\[ (a \times b) \times c = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 36 & 11 & -21 \\ 6 & 2 & -3 \end{vmatrix} \]

Expanding this determinant:

\[ = \hat{i} \begin{vmatrix} 11 & -21 \\ 2 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 36 & -21 \\ 6 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 36 & 11 \\ 6 & 2 \end{vmatrix} \]

Calculating these:

- For \( \hat{i} \):
\[ 11 \cdot (-3) - (-21) \cdot 2 = -33 + 42 = 9 \]

- For \( \hat{j} \):
\[ 36 \cdot (-3) - (-21) \cdot 6 = -108 + 126 = 18 \quad \Rightarrow \quad -18 \]

- For \( \hat{k} \):
\[ 36 \cdot 2 - 11 \cdot 6 = 72 - 66 = 6 \]

Therefore:
\[ (a \times b) \times c = 9 \hat{i} - 18 \hat{j} + 6 \hat{k} \neq 0 \]

Now, calculate \( a \times (b \times c) \):
We already calculated \( b \times c = (14, 21, 42) \).

\[ a \times (b \times c) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ 14 & 21 & 42 \end{vmatrix} \]

Expanding this determinant:

\[ = \hat{i} \begin{vmatrix} 3 & 5 \\ 21 & 42 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 5 \\ 14 & 42 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 14 & 21 \end{vmatrix} \]

Calculating these:

- For \( \hat{i} \):
\[ 3 \cdot 42 - 5 \cdot 21 = 126 - 105 = 21 \]

- For \( \hat{j} \):
\[ 2 \cdot 42 - 5 \cdot 14 = 84 - 70 = 14 \quad \Rightarrow \quad -14 \]

- For \( \hat{k} \):
\[ 2 \cdot 21 - 3 \cdot 14 = 42 - 42 = 0 \]

Therefore:
\[ a \times (b \times c) = 21 \hat{i} - 14 \hat{j} + 0 \hat{k} \neq 0 \]

5. **Conclusion:**
Since both \( (a \times b) \times c \) and \( a \times (b \times c) \) are not equal to the zero vector, the statement \( (a \times b) \times c = a \times (b \times c) = 0 \) is **False**.