If vector `vec i+ 2vec j + 2vec k` is rotated through an angle of `90^@`, so as to cross the positivedirection of y-axis, then the vector in the new position is

Options

A. `-(2)/(sqrt(5))hat(i)+sqrt(5)hat(j)-(4)/(sqrt(5))hat(k)`
B. `-(2)/(sqrt(5))hat(i)-sqrt(5)hat(j)+(4)/(sqrt(5))hat(k)`
C. `4hat(i)-hat(j)-hat(k)`
D. None of these

Answer

To solve the problem of rotating the vector \( \vec{OA} = \hat{i} + 2\hat{j} + 2\hat{k} \) through an angle of \( 90^\circ \) to cross the positive direction of the y-axis, we can follow these steps:

### Step 1: Define the original vector
The original vector is given as:
\[ \vec{OA} = \hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 2: Set up the new vector
Let the new vector after rotation be:
\[ \vec{OB} = x\hat{i} + y\hat{j} + z\hat{k} \]

### Step 3: Equal lengths of vectors
Since the length of the vectors must remain the same after rotation, we have:
\[ |\vec{OA}| = |\vec{OB}| \]
Calculating the magnitude of \( \vec{OA} \):
\[ |\vec{OA}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
Thus, we have:
\[ |\vec{OB}| = \sqrt{x^2 + y^2 + z^2} = 3 \]
Squaring both sides gives:
\[ x^2 + y^2 + z^2 = 9 \quad \text{(Equation 1)} \]

### Step 4: Perpendicularity condition
Since \( \vec{OA} \) is perpendicular to \( \vec{OB} \), their dot product must equal zero:
\[ \vec{OA} \cdot \vec{OB} = 0 \]
Calculating the dot product:
\[ 1 \cdot x + 2 \cdot y + 2 \cdot z = 0 \]
This simplifies to:
\[ x + 2y + 2z = 0 \quad \text{(Equation 2)} \]

### Step 5: Determinant condition
Since \( \vec{OA} \), \( \vec{OB} \), and the y-axis are coplanar, the determinant formed by these vectors must equal zero. The vectors are:
- \( \vec{OA} = (1, 2, 2) \)
- \( \vec{OB} = (x, y, z) \)
- y-axis vector \( (0, \lambda, 0) \)

Setting up the determinant:
\[ \begin{vmatrix} 1 & 2 & 2 \\ x & y & z \\ 0 & \lambda & 0 \end{vmatrix} = 0 \]
Calculating the determinant:
\[ \lambda(2x - 0) - 0 + 0 = 0 \implies 2x = z \quad \text{(Equation 3)} \]

### Step 6: Substitute Equation 3 into Equations 1 and 2
Substituting \( z = 2x \) into Equation 2:
\[ x + 2y + 2(2x) = 0 \implies x + 2y + 4x = 0 \implies 5x + 2y = 0 \implies y = -\frac{5}{2}x \quad \text{(Equation 4)} \]

### Step 7: Substitute Equations 3 and 4 into Equation 1
Substituting \( y = -\frac{5}{2}x \) and \( z = 2x \) into Equation 1:
\[ x^2 + \left(-\frac{5}{2}x\right)^2 + (2x)^2 = 9 \]
Calculating:
\[ x^2 + \frac{25}{4}x^2 + 4x^2 = 9 \]
Combining terms:
\[ x^2 + \frac{25}{4}x^2 + \frac{16}{4}x^2 = 9 \implies \frac{45}{4}x^2 = 9 \]
Multiplying by 4:
\[ 45x^2 = 36 \implies x^2 = \frac{36}{45} = \frac{4}{5} \implies x = \pm \frac{2}{\sqrt{5}} \]

### Step 8: Find values for y and z
Using \( x = \frac{2}{\sqrt{5}} \):
\[ y = -\frac{5}{2} \cdot \frac{2}{\sqrt{5}} = -\frac{5}{\sqrt{5}} = -\sqrt{5} \]
\[ z = 2x = 2 \cdot \frac{2}{\sqrt{5}} = \frac{4}{\sqrt{5}} \]

Using \( x = -\frac{2}{\sqrt{5}} \):
\[ y = -\frac{5}{2} \cdot -\frac{2}{\sqrt{5}} = \sqrt{5} \]
\[ z = 2x = -\frac{4}{\sqrt{5}} \]

### Step 9: Write the new vector
Thus, the two possible vectors \( \vec{OB} \) are:
1. \( \vec{OB} = \frac{2}{\sqrt{5}} \hat{i} - \sqrt{5} \hat{j} + \frac{4}{\sqrt{5}} \hat{k} \)
2. \( \vec{OB} = -\frac{2}{\sqrt{5}} \hat{i} + \sqrt{5} \hat{j} - \frac{4}{\sqrt{5}} \hat{k} \)

Since we need the vector that crosses the positive direction of the y-axis, we take:
\[ \vec{OB} = -\frac{2}{\sqrt{5}} \hat{i} + \sqrt{5} \hat{j} - \frac{4}{\sqrt{5}} \hat{k} \]

### Final Answer
The vector in the new position is:
\[ \vec{OB} = -\frac{2}{\sqrt{5}} \hat{i} + \sqrt{5} \hat{j} - \frac{4}{\sqrt{5}} \hat{k} \]