Given vectors $a = (3, 3, 1)$, $b = (-2, -1, 0)$, and $c = (-2, -3, -1)$, we need to find the following: (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 vectors $a = (3, 3, 1)$, $b = (-2, -1, 0)$, and $c = (-2, -3, -1)$, we need to find the following:

(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) $a \times b$

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

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

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

First, calculate $b + c$:

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

Now, calculate $a \times (b + c)$:

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

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

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

First, calculate $b \times c$:

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

So, $b \times c = (1, -2, 4)$.

Now, calculate $a \cdot (b \times c)$:

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

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

From part (c), we know that $b \times c = (1, -2, 4)$.

Now, calculate $a \times (b \times c)$:

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

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

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