Given three vectors $a = (3, 3, 1)$, $b = (-2, -1, 0)$, and $c = (-2, -3, -1)$, we are asked to find: (a) $a \times b$ (b) $a \times (b + c)$ (c) $a \cdot (b \times c)$ (d) $a \times (b \times c)$

1. Problem Description

Given three vectors $a = (3, 3, 1)$, $b = (-2, -1, 0)$, and $c = (-2, -3, -1)$, we are asked to find:

(a) $a \times b$

(b) $a \times (b + c)$

(c) $a \cdot (b \times c)$

(d) $a \times (b \times c)$

2. Solution Steps

(a) $a \times b = \begin{vmatrix} i & j & k \\ 3 & 3 & 1 \\ -2 & -1 & 0 \end{vmatrix} = (3 \cdot 0 - 1 \cdot (-1))i - (3 \cdot 0 - 1 \cdot (-2))j + (3 \cdot (-1) - 3 \cdot (-2))k = (0 + 1)i - (0 + 2)j + (-3 + 6)k = 1i - 2j + 3k = (1, -2, 3)$

(b) First, we compute $b + c = (-2, -1, 0) + (-2, -3, -1) = (-4, -4, -1)$.

Then, $a \times (b + c) = \begin{vmatrix} i & j & k \\ 3 & 3 & 1 \\ -4 & -4 & -1 \end{vmatrix} = (3 \cdot (-1) - 1 \cdot (-4))i - (3 \cdot (-1) - 1 \cdot (-4))j + (3 \cdot (-4) - 3 \cdot (-4))k = (-3 + 4)i - (-3 + 4)j + (-12 + 12)k = 1i - 1j + 0k = (1, -1, 0)$

(c) First, we compute $b \times c = \begin{vmatrix} i & j & k \\ -2 & -1 & 0 \\ -2 & -3 & -1 \end{vmatrix} = ((-1) \cdot (-1) - 0 \cdot (-3))i - ((-2) \cdot (-1) - 0 \cdot (-2))j + ((-2) \cdot (-3) - (-1) \cdot (-2))k = (1 - 0)i - (2 - 0)j + (6 - 2)k = 1i - 2j + 4k = (1, -2, 4)$.

Then, $a \cdot (b \times c) = (3, 3, 1) \cdot (1, -2, 4) = 3 \cdot 1 + 3 \cdot (-2) + 1 \cdot 4 = 3 - 6 + 4 = 1$.

(d) First, we compute $b \times c = (1, -2, 4)$ from part (c).

Then, $a \times (b \times c) = \begin{vmatrix} i & j & k \\ 3 & 3 & 1 \\ 1 & -2 & 4 \end{vmatrix} = (3 \cdot 4 - 1 \cdot (-2))i - (3 \cdot 4 - 1 \cdot 1)j + (3 \cdot (-2) - 3 \cdot 1)k = (12 + 2)i - (12 - 1)j + (-6 - 3)k = 14i - 11j - 9k = (14, -11, -9)$.

3. Final Answer

(a) $a \times b = (1, -2, 3)$

(b) $a \times (b + c) = (1, -1, 0)$

(c) $a \cdot (b \times c) = 1$

(d) $a \times (b \times c) = (14, -11, -9)$