The angle between two vectors `vec(A)= 3hat(i)+4hat(j)+5hat(k)` and `vec(B)= 3hat(i)+4hat(j)+5hat(k)` is

Options

A. `60^(@)`
B. Zero
C. `90^(@)`
D. None of these

Explanation

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 follow these steps:

### Step 1: Identify the vectors
We have:
\[ \vec{A} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]
\[ \vec{B} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]

### Step 2: Calculate the dot product \(\vec{A} \cdot \vec{B}\)
The dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:
\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z \]
Substituting the components:
\[ \vec{A} \cdot \vec{B} = (3)(3) + (4)(4) + (5)(5) = 9 + 16 + 25 = 50 \]

### Step 3: Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)
The magnitude of a vector \(\vec{A}\) is given by:
\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
Calculating the magnitude of \(\vec{A}\):
\[ |\vec{A}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \]
Since \(\vec{B}\) is the same as \(\vec{A}\), we have:
\[ |\vec{B}| = \sqrt{50} \]

### Step 4: Use the dot product to find the cosine of the angle
The formula relating the dot product and the angle \(\theta\) between two vectors is:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
Substituting the values we calculated:
\[ 50 = \sqrt{50} \cdot \sqrt{50} \cdot \cos \theta \]
This simplifies to:
\[ 50 = 50 \cos \theta \]
Dividing both sides by 50:
\[ \cos \theta = 1 \]

### Step 5: Find the angle \(\theta\)
The angle whose cosine is 1 is:
\[ \theta = \cos^{-1}(1) = 0^\circ \]

### Conclusion
The angle between the two vectors \(\vec{A}\) and \(\vec{B}\) is \(0^\circ\).

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