Step 1: Understanding the Concept:
To find the angle between two vectors $\vec{P}$ and $\vec{Q}$, we use their dot product. If the dot product of two non-zero vectors is zero, they are orthogonal (perpendicular), meaning the angle between them is $90^\circ$.
Step 2: Key Formula or Approach:
1. Vector addition: $\vec{P} = \vec{A} + \vec{B} + \vec{C}$ is found by adding the corresponding $\hat{i}, \hat{j}, \hat{k}$ components.
2. Cross product: $\vec{Q} = \vec{A} \times \vec{B}$ is calculated using the determinant method.
3. Dot product: $\vec{P} \cdot \vec{Q} = |\vec{P}| |\vec{Q}| \cos\theta$. If $\vec{P} \cdot \vec{Q} = 0$, then $\cos\theta = 0 \implies \theta = 90^\circ$.
Step 3: Detailed Explanation:
First, calculate vector $\vec{P}$:
$\vec{P} = (\hat{i} + \hat{j} + 3\hat{k}) + (-\hat{i} + \hat{j} + 4\hat{k}) + (2\hat{i} - 2\hat{j} - 8\hat{k})$
$\vec{P} = (1 - 1 + 2)\hat{i} + (1 + 1 - 2)\hat{j} + (3 + 4 - 8)\hat{k}$
$\vec{P} = 2\hat{i} + 0\hat{j} - 1\hat{k} = 2\hat{i} - \hat{k}$
Next, calculate vector $\vec{Q} = \vec{A} \times \vec{B}$:
\[ \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & 3
-1 & 1 & 4 \end{vmatrix} \]
$\vec{Q} = \hat{i}[(1)(4) - (3)(1)] - \hat{j}[(1)(4) - (3)(-1)] + \hat{k}[(1)(1) - (1)(-1)]$
$\vec{Q} = \hat{i}[4 - 3] - \hat{j}[4 + 3] + \hat{k}[1 + 1]$
$\vec{Q} = 1\hat{i} - 7\hat{j} + 2\hat{k}$
Now, find the dot product $\vec{P} \cdot \vec{Q}$:
$\vec{P} \cdot \vec{Q} = (2\hat{i} + 0\hat{j} - 1\hat{k}) \cdot (1\hat{i} - 7\hat{j} + 2\hat{k})$
$\vec{P} \cdot \vec{Q} = (2)(1) + (0)(-7) + (-1)(2)$
$\vec{P} \cdot \vec{Q} = 2 + 0 - 2 = 0$
Since the dot product is zero and neither $\vec{P}$ nor $\vec{Q}$ is a null vector, the vectors are perpendicular.
Therefore, the angle $\theta$ is $90^\circ$.
Step 4: Final Answer:
The angle between the vectors is $90^\circ$.