The angle between the two Vectors `vec(A)= 3hat(i)+4hat(j)+5hat(k)` and `vec(B)= 3hat(i)+4hat(j)-5hat(k)` will be

Options

A. zero
B. `180^(@)`
C. `90^(@)`
D. `45^(@)`

Answer

To find the angle between the two vectors \(\vec{A} = 3\hat{i} + 4\hat{j} + 5\hat{k}\) and \(\vec{B} = 3\hat{i} + 4\hat{j} - 5\hat{k}\), we can use the formula for the dot product of two vectors, which relates the dot product to the angle between them.

### Step-by-Step Solution:

1. **Calculate the Dot Product of \(\vec{A}\) and \(\vec{B}\)**:
\[ \vec{A} \cdot \vec{B} = (3\hat{i} + 4\hat{j} + 5\hat{k}) \cdot (3\hat{i} + 4\hat{j} - 5\hat{k}) \]
Using the properties of dot product:
\[ = 3 \cdot 3 + 4 \cdot 4 + 5 \cdot (-5) \]
\[ = 9 + 16 - 25 = 0 \]

2. **Calculate the Magnitudes of \(\vec{A}\) and \(\vec{B}\)**:
- For \(\vec{A}\):
\[ |\vec{A}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \]
- For \(\vec{B}\):
\[ |\vec{B}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \]

3. **Use the Dot Product to Find the Cosine of the Angle**:
The formula relating the dot product to the angle is:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \]
Substituting the values we found:
\[ 0 = |\vec{A}| |\vec{B}| \cos(\theta) \]
Since \(|\vec{A}|\) and \(|\vec{B}|\) are both non-zero, we can conclude:
\[ \cos(\theta) = 0 \]

4. **Determine the Angle \(\theta\)**:
The angle \(\theta\) for which \(\cos(\theta) = 0\) is:
\[ \theta = 90^\circ \]

### Final Answer:
The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).

---