Given three vectors $\vec{a} = 6\hat{i} + 3\hat{j} - 9\hat{k}$, $\vec{b} = 12\hat{i} - 8\hat{j} - 4\hat{k}$, and $\vec{c} = 4\hat{i} - 2\hat{j} + 3\hat{k}$, we need to find: I. The angle between $\vec{a}$ and $\vec{b}$. II. The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$. III. The vector triple product $\vec{c} \times (\vec{a} \times \vec{b})$.

1. Problem Description

Given three vectors $\vec{a} = 6\hat{i} + 3\hat{j} - 9\hat{k}$, $\vec{b} = 12\hat{i} - 8\hat{j} - 4\hat{k}$, and $\vec{c} = 4\hat{i} - 2\hat{j} + 3\hat{k}$, we need to find:

I. The angle between $\vec{a}$ and $\vec{b}$.

II. The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$.

III. The vector triple product $\vec{c} \times (\vec{a} \times \vec{b})$.

2. Solution Steps

I. Angle between $\vec{a}$ and $\vec{b}$:

The dot product of two vectors is given by:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}$

where $\theta$ is the angle between the two vectors. Therefore,

$\cos{\theta} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

$\vec{a} \cdot \vec{b} = (6)(12) + (3)(-8) + (-9)(-4) = 72 - 24 + 36 = 84$

$|\vec{a}| = \sqrt{6^2 + 3^2 + (-9)^2} = \sqrt{36 + 9 + 81} = \sqrt{126} = 3\sqrt{14}$

$|\vec{b}| = \sqrt{12^2 + (-8)^2 + (-4)^2} = \sqrt{144 + 64 + 16} = \sqrt{224} = 4\sqrt{14}$

$\cos{\theta} = \frac{84}{(3\sqrt{14})(4\sqrt{14})} = \frac{84}{12 \cdot 14} = \frac{84}{168} = \frac{1}{2}$

$\theta = \arccos{\frac{1}{2}} = \frac{\pi}{3}$ radians or $60^{\circ}$

II. Evaluate $\vec{a} \cdot (\vec{b} \times \vec{c})$:

The scalar triple product can be calculated as the determinant of a matrix formed by the components of the three vectors:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix} = \begin{vmatrix} 6 & 3 & -9 \\ 12 & -8 & -4 \\ 4 & -2 & 3 \end{vmatrix}$

Expanding the determinant:

$= 6((-8)(3) - (-4)(-2)) - 3((12)(3) - (-4)(4)) + (-9)((12)(-2) - (-8)(4))$

$= 6(-24 - 8) - 3(36 + 16) - 9(-24 + 32)$

$= 6(-32) - 3(52) - 9(8)$

$= -192 - 156 - 72$

$= -420$

III. Evaluate $\vec{c} \times (\vec{a} \times \vec{b})$:

First, we calculate $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 3 & -9 \\ 12 & -8 & -4 \end{vmatrix} = \hat{i}((3)(-4) - (-9)(-8)) - \hat{j}((6)(-4) - (-9)(12)) + \hat{k}((6)(-8) - (3)(12))$

$= \hat{i}(-12 - 72) - \hat{j}(-24 + 108) + \hat{k}(-48 - 36)$

$= -84\hat{i} - 84\hat{j} - 84\hat{k}$

Now, we calculate $\vec{c} \times (\vec{a} \times \vec{b})$:

$\vec{c} \times (\vec{a} \times \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 3 \\ -84 & -84 & -84 \end{vmatrix} = \hat{i}((-2)(-84) - (3)(-84)) - \hat{j}((4)(-84) - (3)(-84)) + \hat{k}((4)(-84) - (-2)(-84))$

$= \hat{i}(168 + 252) - \hat{j}(-336 + 252) + \hat{k}(-336 - 168)$

$= 420\hat{i} + 84\hat{j} - 504\hat{k}$

3. Final Answer

I. The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$ radians or $60^{\circ}$.

II. $\vec{a} \cdot (\vec{b} \times \vec{c}) = -420$

III. $\vec{c} \times (\vec{a} \times \vec{b}) = 420\hat{i} + 84\hat{j} - 504\hat{k}$