Find the scalar product of the following pairs of vectors and the angle between them :
`2hat(i)-3hat(j)+6hat(k) " and " 2hat(i)-3hat(j)-5hat(k)`

Answer

To find the scalar product (dot product) of the vectors \( \mathbf{A} = 2\hat{i} - 3\hat{j} + 6\hat{k} \) and \( \mathbf{B} = 2\hat{i} - 3\hat{j} - 5\hat{k} \), and the angle between them, we can follow these steps:

### Step 1: Calculate the Scalar Product (Dot Product)

The scalar 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 \( \mathbf{A} \)
- \( B_x, B_y, B_z \) are the components of vector \( \mathbf{B} \)

For our vectors:
- \( \mathbf{A} = (2, -3, 6) \)
- \( \mathbf{B} = (2, -3, -5) \)

Calculating the dot product:

\[ \mathbf{A} \cdot \mathbf{B} = (2)(2) + (-3)(-3) + (6)(-5) \]

Calculating each term:

1. \( (2)(2) = 4 \)
2. \( (-3)(-3) = 9 \)
3. \( (6)(-5) = -30 \)

Now, summing these results:

\[ \mathbf{A} \cdot \mathbf{B} = 4 + 9 - 30 = 13 - 30 = -17 \]

### Step 2: 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 \( \mathbf{A} \):

\[ |\mathbf{A}| = \sqrt{(2)^2 + (-3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

Now, calculating the magnitude of \( \mathbf{B} \):

\[ |\mathbf{B}| = \sqrt{(2)^2 + (-3)^2 + (-5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38} \]

### Step 3: Calculate the Angle Between the Vectors

The angle \( \theta \) between two vectors can be found using the formula:

\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

Substituting the values we found:

\[ \cos \theta = \frac{-17}{7 \sqrt{38}} \]

### Step 4: Find the Angle \( \theta \)

To find \( \theta \), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{-17}{7 \sqrt{38}}\right) \]

Since the cosine value is negative, \( \theta \) will be in the second quadrant.

### Final Answer

The scalar product of the vectors is \( -17 \) and the angle between them is:

\[ \theta = \pi - \cos^{-1}\left(\frac{17}{7 \sqrt{38}}\right) \]