What is the angle between `vecA = 5hati - 5hatj and vecB = 5hati-5hatj` ?

Options

A. `90^@`
B. `45^@`
C. `0^@`
D. `60^@`

Explanation

To find the angle between the vectors \( \vec{A} = 5 \hat{i} - 5 \hat{j} \) and \( \vec{B} = 5 \hat{i} - 5 \hat{j} \), we can follow these steps:

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

### 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:
\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y \]
For our vectors:
\[ \vec{A} \cdot \vec{B} = (5)(5) + (-5)(-5) = 25 + 25 = 50 \]

### Step 3: Calculate the magnitudes of \( \vec{A} \) and \( \vec{B} \)
The magnitude of a vector \( \vec{A} = a \hat{i} + b \hat{j} \) is given by:
\[ |\vec{A}| = \sqrt{a^2 + b^2} \]
For \( \vec{A} \):
\[ |\vec{A}| = \sqrt{(5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} \]
Similarly, for \( \vec{B} \):
\[ |\vec{B}| = \sqrt{(5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} \]

### Step 4: Use the dot product to find the cosine of the angle
The cosine of the angle \( \theta \) between two vectors can be found using the formula:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{50}{\sqrt{50} \cdot \sqrt{50}} = \frac{50}{50} = 1 \]

### Step 5: Find the angle \( \theta \)
Since \( \cos \theta = 1 \), we find:
\[ \theta = \cos^{-1}(1) = 0^\circ \]

### Final Answer
The angle between \( \vec{A} \) and \( \vec{B} \) is \( 0^\circ \).

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