Let a, b be two vectors perependicular to each other and `|a|=2, |b|=3 and ctimesa=b.` Q. When |c-a| is least the value of `alpha` (when `alpha` is angle between a and c) equals

Options

A. `tan^(-1)(2)`
B. `tan^(-1)(3)/(4)`
C. `cos^(-1)((2)/(3))`
D. None of these

Answer

To solve the problem step by step, we will analyze the given information and use vector algebra to find the angle \( \alpha \) between vectors \( \mathbf{a} \) and \( \mathbf{c} \) when \( |\mathbf{c} - \mathbf{a}| \) is minimized.

### Step 1: Understand the Given Information
We are given:
- Vectors \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular.
- Magnitudes: \( |\mathbf{a}| = 2 \) and \( |\mathbf{b}| = 3 \).
- The relationship \( \mathbf{c} \times \mathbf{a} = \mathbf{b} \).

### Step 2: Use the Cross Product
From the equation \( \mathbf{c} \times \mathbf{a} = \mathbf{b} \), we can take the dot product of both sides with \( \mathbf{a} \):
\[ \mathbf{a} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{a} \cdot \mathbf{b} \]
Since \( \mathbf{a} \times \mathbf{a} = \mathbf{0} \), the left side becomes zero:
\[ 0 = \mathbf{a} \cdot \mathbf{b} \]
This confirms that \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular.

### Step 3: Vector Triple Product Identity
Using the vector triple product identity:
\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \]
We can rewrite \( \mathbf{c} \times \mathbf{a} = \mathbf{b} \) as:
\[ \mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{a} \]

### Step 4: Magnitudes and Dot Products
We know:
- \( |\mathbf{a}|^2 = 4 \)
- \( |\mathbf{b}|^2 = 9 \)

From the equation \( \mathbf{c} \times \mathbf{a} = \mathbf{b} \), we can find the magnitude:
\[ |\mathbf{c} \times \mathbf{a}| = |\mathbf{b}| = 3 \]
Using the formula for the magnitude of the cross product:
\[ |\mathbf{c}||\mathbf{a}|\sin \alpha = 3 \]

### Step 5: Express \( |\mathbf{c}| \)
Let \( |\mathbf{c}| = c \). Then:
\[ c \cdot 2 \cdot \sin \alpha = 3 \implies c \sin \alpha = \frac{3}{2} \]

### Step 6: Minimize \( |\mathbf{c} - \mathbf{a}| \)
We want to minimize \( |\mathbf{c} - \mathbf{a}| \):
\[ |\mathbf{c} - \mathbf{a}|^2 = |\mathbf{c}|^2 + |\mathbf{a}|^2 - 2|\mathbf{c}||\mathbf{a}|\cos \alpha \]
Substituting \( |\mathbf{a}| = 2 \):
\[ |\mathbf{c} - \mathbf{a}|^2 = c^2 + 4 - 4c \cos \alpha \]

### Step 7: Substitute \( c \sin \alpha \)
From \( c \sin \alpha = \frac{3}{2} \), we can express \( c \) in terms of \( \sin \alpha \):
\[ c = \frac{3}{2 \sin \alpha} \]

### Step 8: Substitute into the Expression
Substituting \( c \) into the equation:
\[ |\mathbf{c} - \mathbf{a}|^2 = \left(\frac{3}{2 \sin \alpha}\right)^2 + 4 - 4 \left(\frac{3}{2 \sin \alpha}\right)(2 \cos \alpha) \]

### Step 9: Find the Minimum
To minimize this expression, we can differentiate with respect to \( \alpha \) or use the condition for minimum value of quadratic equations. The minimum occurs when:
\[ \tan \alpha = \frac{3}{4} \]
Thus:
\[ \alpha = \tan^{-1}\left(\frac{3}{4}\right) \]

### Final Answer
The value of \( \alpha \) when \( |\mathbf{c} - \mathbf{a}| \) is minimized is:
\[ \alpha = \tan^{-1}\left(\frac{3}{4}\right) \]