Find the angle between the vectors `(-2i + 2j) and 3i`

Options

A. `60^(@)`
B. `120^(@)`
C. `135^(@)`
D. `45^(@)`

Answer

To find the angle between the vectors \((-2\hat{i} + 2\hat{j})\) and \(3\hat{i}\), we can follow these steps:

### Step 1: Identify the vectors
Let vector **A** be \(\mathbf{A} = -2\hat{i} + 2\hat{j}\) and vector **B** be \(\mathbf{B} = 3\hat{i}\).

### Step 2: Use the formula for the angle between two vectors
The angle \(\theta\) between two vectors can be calculated using the formula:
\[ \cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \]

### Step 3: Calculate the dot product \(\mathbf{A} \cdot \mathbf{B}\)
The dot product of vectors \(\mathbf{A}\) and \(\mathbf{B}\) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (-2\hat{i} + 2\hat{j}) \cdot (3\hat{i}) = (-2 \cdot 3) + (2 \cdot 0) = -6 + 0 = -6 \]

### Step 4: Calculate the magnitudes of the vectors
Now, we need to find the magnitudes of vectors \(\mathbf{A}\) and \(\mathbf{B}\):
\[ |\mathbf{A}| = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
\[ |\mathbf{B}| = \sqrt{(3)^2} = \sqrt{9} = 3 \]

### Step 5: Substitute the values into the formula
Now substitute the values into the cosine formula:
\[ \cos(\theta) = \frac{-6}{(2\sqrt{2})(3)} = \frac{-6}{6\sqrt{2}} = \frac{-1}{\sqrt{2}} \]

### Step 6: Find the angle \(\theta\)
To find \(\theta\), take the inverse cosine:
\[ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \]
The angle corresponding to \(-\frac{1}{\sqrt{2}}\) is \(135^\circ\).

### Final Answer
Thus, the angle between the vectors \((-2\hat{i} + 2\hat{j})\) and \(3\hat{i}\) is:
\[ \theta = 135^\circ \]