Given vectors a and b, find the unit vector in the direction of a and the angle between a and b.

1a)

Step 1: Find the magnitude of vector $a$.
Given $a = 3i - 4j + 5k$, its components are $(3, -4, 5)$.
The magnitude $|a|$ is given by:

$$|a| = \sqrt{3^2 + (-4)^2 + 5^2}$$ $$|a| = \sqrt{9 + 16 + 25}$$ $$|a| = \sqrt{50}$$ $$|a| = 5\sqrt{2}$$

Step 2: Find the unit vector in the direction of $a$.
The unit vector $\hat{a}$ is given by $\frac{a}{|a|}$.

$$\hat{a} = \frac{1}{5\sqrt{2}}(3i - 4j + 5k)$$ $$\hat{a} = \frac{3}{5\sqrt{2}}i - \frac{4}{5\sqrt{2}}j + \frac{5}{5\sqrt{2}}k$$

Rationalize the denominators:

$$\hat{a} = \frac{3\sqrt{2}}{10}i - \frac{4\sqrt{2}}{10}j + \frac{5\sqrt{2}}{10}k$$ $$\hat{a} = \frac{3\sqrt{2}}{10}i - \frac{2\sqrt{2}}{5}j + \frac{\sqrt{2}}{2}k$$

The unit vector is $\boxed{\frac{3\sqrt{2}}{10}i - \frac{2\sqrt{2}}{5}j + \frac{\sqrt{2}}{2}k}$.

1b)

Step 1: Compute the dot product $a \cdot b$.
Given $a = 2i - j + 3k$ and $b = i + 4j - 2k$.

$$a \cdot b = (2)(1) + (-1)(4) + (3)(-2)$$ $$a \cdot b = 2 - 4 - 6$$ $$a \cdot b = -8$$

Step 2: Compute the magnitudes of $a$ and $b$.

$$|a| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$$ $$|b| = \sqrt{1^2 + 4^2 + (-2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21}$$

Step 3: Compute the angle $\theta$ between $a$ and $b$.
Using the formula $\cos \theta = \frac{a \cdot b}{|a||b|}$:

$$\cos \theta = \frac{-8}{\sqrt{14}\sqrt{21}}$$ $$\cos \theta = \frac{-8}{\sqrt{294}}$$ $$\cos \theta = \frac{-8}{7\sqrt{6}}$$

Rationalize the denominator:

$$\cos \theta = \frac{-8\sqrt{6}}{42}$$ $$\cos \theta = \frac{-4\sqrt{6}}{21}$$ $$\theta = \arccos\left(\frac{-4\sqrt{6}}{21}\right)$$ $$\theta \approx \arccos(-0.4666)$$ $$\theta \approx 117.8^\circ$$

The dot product is $\boxed{-8}$ and the angle between them is approximately $\boxed{117.8^\circ}$.

1c)

Step 1: Identify vectors $a$ and $b$.
Given $a = i + 2j - k$ and $b = 3i - j + 2k$.

Step 2: Compute the dot product $a \cdot b$.

$$a \cdot b = (1)(3) + (2)(-1) + (-1)(2)$$ $$a \cdot b = 3 - 2 - 2$$ $$a \cdot b = -1$$

Step 3: Compute the magnitude squared of $b$, $|b|^2$.

$$|b|^2 = 3^2 + (-1)^2 + 2^2$$ $$|b|^2 = 9 + 1 + 4$$ $$|b|^2 = 14$$

Step 4: Compute the vector projection of $a$ onto $b$.
The formula for the vector projection of $a$ onto $b$ is $\text{proj}_b a = \frac{a \cdot b}{|b|^2} b$.

$$\text{proj}_b a = \frac{-1}{14}(3i - j + 2k)$$ $$\text{proj}_b a = -\frac{3}{14}i + \frac{1}{14}j - \frac{2}{14}k$$ $$\text{proj}_b a = -\frac{3}{14}i + \frac{1}{14}j - \frac{1}{7}k$$

The vector projection is $\boxed{-\frac{3}{14}i + \frac{1}{14}j - \frac{1}{7}k}$.

2a)

Step 1: Identify vectors $u$ and $v$.
Given $u = (1, -2, 3)$ and $v = (4, 0, -1)$.

Step 2: Compute the cross product $u \times v$.

$$u \times v = \begin{vmatrix} i & j & k \\ 1 & -2 & 3 \\ 4 & 0 & -1 \end{vmatrix}$$ $$u \times v = i((-2)(-1) - (3)(0)) - j((1)(-1) - (3)(4)) + k((1)(0) - (-2)(4))$$ $$u \times v = i(2 - 0) - j(-1 - 12) + k(0 - (-8))$$ $$u \times v = 2i - j(-13) + k(8)$$ $$u \times v = 2i + 13j + 8k$$

The cross product is $\boxed{2i + 13j + 8k}$.

2b)

Step 1: Identify vectors $p$ and $q$.
Given $p = 2i + j - k$ and $q = i - 3j + 4k$.

Step 2: Compute the cross product $p \times q$.

$$p \times q = \begin{vmatrix} i & j & k \\ 2 & 1 & -1 \\ 1 & -3 & 4 \end{vmatrix}$$ $$p \times q = i((1)(4) - (-1)(-3)) - j((2)(4) - (-1)(1)) + k((2)(-3) - (1)(1))$$ $$p \times q = i(4 - 3) - j(8 - (-1)) + k(-6 - 1)$$ $$p \times q = i(1) - j(9) + k(-7)$$ $$p \times q = i - 9j - 7k$$

Step 3: Find the magnitude of the cross product.
The area of the parallelogram is $|p \times q|$.

$$|p \times q| = \sqrt{1^2 + (-9)^2 + (-7)^2}$$ $$|p \times q| = \sqrt{1 + 81 + 49}$$ $$|p \times q| = \sqrt{131}$$

The area of the parallelogram is $\boxed{\sqrt{131}}$ square units.

2c)

Step 1: Form a matrix with the given vectors as rows.
Given $a = (1, 2, 3)$, $b = (4, 5, 6)$, and $c = (7, 8, 9)$.
Let $M = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$.

Step 2: Calculate the determinant of the matrix $M$.
If $\det(M) = 0$, the vectors are linearly dependent. If $\det(M) \neq 0$, they are linearly independent.

$$\det(M) = 1(5 \cdot 9 - 6 \cdot 8) - 2(4 \cdot 9 - 6 \cdot 7) + 3(4 \cdot 8 - 5 \cdot 7)$$ $$\det(M) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35)$$ $$\det(M) = 1(-3) - 2(-6) + 3(-3)$$ $$\det(M) = -3 + 12 - 9$$ $$\det(M) = 0$$

Since the determinant is 0, the vectors are linearly dependent.
The vectors are $\boxed{\text{linearly dependent}}$.

3a)

Step 1: Compute $AB$.

$$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}, B = \begin{pmatrix} 2 & 0 & 1 \\ -1 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}$$ $$AB = \begin{pmatrix} (1)(2)+(2)(-1)+(3)(3) & (1)(0)+(2)(1)+(3)(2) & (1)(1)+(2)(0)+(3)(1) \\ (0)(2)+(1)(-1)+(4)(3) & (0)(0)+(1)(1)+(4)(2) & (0)(1)+(1)(0)+(4)(1) \\ (5)(2)+(6)(-1)+(0)(3) & (5)(0)+(6)(1)+(0)(2) & (5)(1)+(6)(0)+(0)(1) \end{pmatrix}$$ $$AB = \begin{pmatrix} 2-2+9 & 0+2+6 & 1+0+3 \\ 0-1+12 & 0+1+8 & 0+0+4 \\ 10-6+0 & 0+6+0 & 5+0+0 \end{pmatrix}$$ $$AB = \begin{pmatrix} 9 & 8 & 4 \\ 11 & 9 & 4 \\ 4 & 6 & 5 \end{pmatrix}$$

Step 2: Compute $BA$.

$$BA = \begin{pmatrix} 2 & 0 & 1 \\ -1 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$$ $$BA = \begin{pmatrix} (2)(1)+(0)(0)+(1)(5) & (2)(2)+(0)(1)+(1)(6) & (2)(3)+(0)(4)+(1)(0) \\ (-1)(1)+(1)(0)+(0)(5) & (-1)(2)+(1)(1)+(0)(6) & (-1)(3)+(1)(4)+(0)(0) \\ (3)(1)+(2)(0)+(1)(5) & (3)(2)+(2)(1)+(1)(6) & (3)(3)+(2)(4)+(1)(0) \end{pmatrix}$$ $$BA = \begin{pmatrix} 2+0+5 & 4+0+6 & 6+0+0 \\ -1+0+0 & -2+1+0 & -3+4+0 \\ 3+0+5 & 6+2+6 & 9+8+0 \end{pmatrix}$$ $$BA = \begin{pmatrix} 7 & 10 & 6 \\ -1 & -1 & 1 \\ 8 & 14 & 17 \end{pmatrix}$$

Step 3: Compute $AB - BA$.

$$AB - BA = \begin{pmatrix} 9 & 8 & 4 \\ 11 & 9 & 4 \\ 4 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 7 & 10 & 6 \\ -1 & -1 & 1 \\ 8 & 14 & 17 \end{pmatrix}$$ $$AB - BA = \begin{pmatrix} 9-7 & 8-10 & 4-6 \\ 11-(-1) & 9-(-1) & 4-1 \\ 4-8 & 6-14 & 5-17 \end{pmatrix}$$ $$AB - BA = \begin{pmatrix} 2 & -2 & -2 \\ 12 & 10 & 3 \\ -4 & -8 & -12 \end{pmatrix}$$

The result is $\boxed{\begin{pmatrix} 2 & -2 & -2 \\ 12 & 10 & 3 \\ -4 & -8 & -12 \end{pmatrix}}$.

3b)

Step 1: Identify matrix $C$.

$$$$