We are given three vectors, $a = -3i + 2j - 2k$, $b = -i + 2j - 4k$, and $c = 7i + 3j - 4k$. We need to find: (a) $a \times b$ (b) $a \times (b + c)$ (c) $a \cdot (b + c)$ (d) $a \times (b \times c)$

1. Problem Description

We are given three vectors, $a = -3i + 2j - 2k$, $b = -i + 2j - 4k$, and $c = 7i + 3j - 4k$. We need to find:

(a) $a \times b$

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

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

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

2. Solution Steps

(a) $a \times b$:

$a \times b = \begin{vmatrix} i & j & k \\ -3 & 2 & -2 \\ -1 & 2 & -4 \end{vmatrix} = i(2(-4) - (-2)(2)) - j((-3)(-4) - (-2)(-1)) + k((-3)(2) - (2)(-1)) \\ = i(-8 + 4) - j(12 - 2) + k(-6 + 2) = -4i - 10j - 4k$

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

First, find $b + c = (-1 + 7)i + (2 + 3)j + (-4 + (-4))k = 6i + 5j - 8k$.

Then, $a \times (b + c) = \begin{vmatrix} i & j & k \\ -3 & 2 & -2 \\ 6 & 5 & -8 \end{vmatrix} = i(2(-8) - (-2)(5)) - j((-3)(-8) - (-2)(6)) + k((-3)(5) - (2)(6)) \\ = i(-16 + 10) - j(24 + 12) + k(-15 - 12) = -6i - 36j - 27k$

(c) $a \cdot (b + c)$:

We already found $b + c = 6i + 5j - 8k$.

$a \cdot (b + c) = (-3)(6) + (2)(5) + (-2)(-8) = -18 + 10 + 16 = 8$

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

First, find $b \times c = \begin{vmatrix} i & j & k \\ -1 & 2 & -4 \\ 7 & 3 & -4 \end{vmatrix} = i(2(-4) - (-4)(3)) - j((-1)(-4) - (-4)(7)) + k((-1)(3) - (2)(7)) \\ = i(-8 + 12) - j(4 + 28) + k(-3 - 14) = 4i - 32j - 17k$

Then, $a \times (b \times c) = \begin{vmatrix} i & j & k \\ -3 & 2 & -2 \\ 4 & -32 & -17 \end{vmatrix} = i(2(-17) - (-2)(-32)) - j((-3)(-17) - (-2)(4)) + k((-3)(-32) - (2)(4)) \\ = i(-34 - 64) - j(51 + 8) + k(96 - 8) = -98i - 59j + 88k$

3. Final Answer

(a) $a \times b = -4i - 10j - 4k$

(b) $a \times (b + c) = -6i - 36j - 27k$

(c) $a \cdot (b + c) = 8$

(d) $a \times (b \times c) = -98i - 59j + 88k$