Find the scalar product of the following pairs of vectors and the angle between them :
`hat(i)+3hat(j)-8hat(k) " and " -3hat(i)-5hat(j)+4hat(k)`.

Answer

To solve the problem of finding the scalar product and the angle between the vectors \( \mathbf{A} = \hat{i} + 3\hat{j} - 8\hat{k} \) and \( \mathbf{B} = -3\hat{i} - 5\hat{j} + 4\hat{k} \), we will follow these steps:

### Step 1: Identify the vectors
Let:
\[ \mathbf{A} = \hat{i} + 3\hat{j} - 8\hat{k} \]
\[ \mathbf{B} = -3\hat{i} - 5\hat{j} + 4\hat{k} \]

### Step 2: Calculate the scalar (dot) product
The scalar product (dot product) of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \]
where \( A_x, A_y, A_z \) are the components of vector \( \mathbf{A} \) and \( B_x, B_y, B_z \) are the components of vector \( \mathbf{B} \).

Calculating the components:
- \( A_x = 1, A_y = 3, A_z = -8 \)
- \( B_x = -3, B_y = -5, B_z = 4 \)

Now, substituting these values:
\[ \mathbf{A} \cdot \mathbf{B} = (1)(-3) + (3)(-5) + (-8)(4) \]
Calculating each term:
\[ = -3 - 15 - 32 \]
Adding these together:
\[ = -50 \]

### Step 3: Calculate the magnitudes of the vectors
The magnitude of a vector \( \mathbf{A} \) is given by:
\[ |\mathbf{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
Calculating the magnitude of \( \mathbf{A} \):
\[ |\mathbf{A}| = \sqrt{1^2 + 3^2 + (-8)^2} = \sqrt{1 + 9 + 64} = \sqrt{74} \]

Calculating the magnitude of \( \mathbf{B} \):
\[ |\mathbf{B}| = \sqrt{(-3)^2 + (-5)^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} \]

### Step 4: Use the dot product to find the angle
The relationship between the dot product and the angle \( \theta \) between two vectors is given by:
\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos \theta \]
Substituting the known values:
\[ -50 = \sqrt{74} \cdot \sqrt{50} \cdot \cos \theta \]
\[ \cos \theta = \frac{-50}{\sqrt{74} \cdot \sqrt{50}} \]

### Step 5: Calculate \( \theta \)
To find \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{-50}{\sqrt{74 \cdot 50}}\right) \]

Calculating \( \sqrt{74 \cdot 50} \):
\[ \sqrt{74 \cdot 50} = \sqrt{3700} = 10\sqrt{37} \]
Thus,
\[ \cos \theta = \frac{-50}{10\sqrt{37}} = \frac{-5}{\sqrt{37}} \]
Now, substituting back to find \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{-5}{\sqrt{37}}\right) \]

### Final Answers
1. The scalar product \( \mathbf{A} \cdot \mathbf{B} = -50 \).
2. The angle \( \theta = \cos^{-1}\left(\frac{-5}{\sqrt{37}}\right) \).