The vector `vec(OA)` where O is origin is given by `vec(OA) = 2 hati + 2 hatj` . Now it is rotated by `45^(@)` anticlockwise about O . What will be the new vector . ?

Options

A. `2 sqrt(2) hatj`
B. `2 hatj`
C. `2 hatj`
D. `2sqrt(2) hatj`

Answer

To find the new vector after rotating the vector \(\vec{OA} = 2 \hat{i} + 2 \hat{j}\) by \(45^\circ\) anticlockwise about the origin, we can follow these steps:

### Step 1: Understand the Initial Vector
The initial vector \(\vec{OA}\) can be represented in Cartesian coordinates as follows:
\[ \vec{OA} = 2 \hat{i} + 2 \hat{j} \]
This means the vector has a component of \(2\) units in the \(x\)-direction and \(2\) units in the \(y\)-direction.

### Step 2: Determine the Angle of the Vector
To find the angle \(\theta\) that the vector makes with the positive \(x\)-axis, we can use the tangent function:
\[ \tan(\theta) = \frac{y}{x} = \frac{2}{2} = 1 \]
This gives us:
\[ \theta = 45^\circ \]

### Step 3: Apply the Rotation Matrix
When rotating a vector in the plane, we can use the rotation matrix for an angle \(\theta\):
\[ R(\theta) = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \]
For a \(45^\circ\) rotation, we have:
\[ R(45^\circ) = \begin{pmatrix} \cos(45^\circ) & -\sin(45^\circ) \\ \sin(45^\circ) & \cos(45^\circ) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix} \]

### Step 4: Represent the Vector as a Column Matrix
We can represent the vector \(\vec{OA}\) as a column matrix:
\[ \begin{pmatrix} 2 \\ 2 \end{pmatrix} \]

### Step 5: Multiply the Rotation Matrix by the Vector
Now we multiply the rotation matrix by the vector:
\[ \begin{pmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix} \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{2}}{2} \cdot 2 - \frac{\sqrt{2}}{2} \cdot 2 \\ \frac{\sqrt{2}}{2} \cdot 2 + \frac{\sqrt{2}}{2} \cdot 2 \end{pmatrix} \]

Calculating this gives:
\[ = \begin{pmatrix} 0 \\ 2\sqrt{2} \end{pmatrix} \]

### Step 6: Write the New Vector
Thus, the new vector after rotation is:
\[ \vec{OA'} = 0 \hat{i} + 2\sqrt{2} \hat{j} \]

### Final Answer
The new vector after rotating \(\vec{OA}\) by \(45^\circ\) anticlockwise about the origin is:
\[ \vec{OA'} = 0 \hat{i} + 2\sqrt{2} \hat{j} \]

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