The angle between `vecA = hati + hatj and vecB = hati - hatj` is

Options

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

Explanation

To find the angle between the vectors \(\vec{A} = \hat{i} + \hat{j}\) and \(\vec{B} = \hat{i} - \hat{j}\), we can use the formula for the cosine of the angle \(\theta\) between two vectors:

\[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \]

**Step 1: Calculate the dot product \(\vec{A} \cdot \vec{B}\)**

The dot product of two vectors \(\vec{A} = \hat{i} + \hat{j}\) and \(\vec{B} = \hat{i} - \hat{j}\) is calculated as follows:

\[ \vec{A} \cdot \vec{B} = (\hat{i} + \hat{j}) \cdot (\hat{i} - \hat{j}) = \hat{i} \cdot \hat{i} + \hat{j} \cdot (-\hat{j}) \]

Calculating the individual components:

\[ \hat{i} \cdot \hat{i} = 1 \quad \text{and} \quad \hat{j} \cdot (-\hat{j}) = -1 \]

Thus,

\[ \vec{A} \cdot \vec{B} = 1 - 1 = 0 \]

**Step 2: Calculate the magnitudes of \(\vec{A}\) and \(\vec{B}\)**

The magnitude of \(\vec{A}\) is given by:

\[ |\vec{A}| = \sqrt{(\text{coefficient of } \hat{i})^2 + (\text{coefficient of } \hat{j})^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

The magnitude of \(\vec{B}\) is given by:

\[ |\vec{B}| = \sqrt{(\text{coefficient of } \hat{i})^2 + (\text{coefficient of } \hat{j})^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]

**Step 3: Substitute values into the cosine formula**

Now substituting the values into the cosine formula:

\[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{0}{\sqrt{2} \cdot \sqrt{2}} = \frac{0}{2} = 0 \]

**Step 4: Determine the angle \(\theta\)**

Since \(\cos(\theta) = 0\), this implies:

\[ \theta = 90^\circ \]

Thus, the angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).

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

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