If the magnitude of the projection vector of the vector `alphahati+betahatj on sqrt(3) hati+hatj is sqrt(3)` and if `alpha=2+sqrt(3)beta` then possible value (s) of `|alpha|` is /are

Options

A. 1,2
B. 3,4
C. 4,5
D. 3

Explanation

To solve the problem step by step, we will follow the given information and apply the projection formula.

### Step 1: Understand the projection formula
The magnitude of the projection of vector **A** on vector **B** is given by:
\[ \text{Magnitude of projection} = \frac{|\mathbf{A} \cdot \mathbf{B}|}{|\mathbf{B}|} \]
where \(\mathbf{A} = \alpha \hat{i} + \beta \hat{j}\) and \(\mathbf{B} = \sqrt{3} \hat{i} + \hat{j}\).

### Step 2: Calculate the dot product \(\mathbf{A} \cdot \mathbf{B}\)
Calculating the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = (\alpha \hat{i} + \beta \hat{j}) \cdot (\sqrt{3} \hat{i} + \hat{j}) = \alpha \sqrt{3} + \beta \]

### Step 3: Calculate the magnitude of vector \(\mathbf{B}\)
The magnitude of vector \(\mathbf{B}\) is:
\[ |\mathbf{B}| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

### Step 4: Set up the equation for the projection
Given that the magnitude of the projection is \(\sqrt{3}\), we can set up the equation:
\[ \frac{|\alpha \sqrt{3} + \beta|}{2} = \sqrt{3} \]

### Step 5: Solve for \(|\alpha \sqrt{3} + \beta|\)
Multiplying both sides by 2 gives:
\[ |\alpha \sqrt{3} + \beta| = 2\sqrt{3} \]

### Step 6: Consider the two cases for the absolute value
This leads to two cases:
1. \(\alpha \sqrt{3} + \beta = 2\sqrt{3}\)
2. \(\alpha \sqrt{3} + \beta = -2\sqrt{3}\)

### Step 7: Substitute \(\alpha\) in terms of \(\beta\)
We are given that \(\alpha = 2 + \sqrt{3} \beta\). Substitute this into both cases.

#### Case 1:
\[ (2 + \sqrt{3} \beta) \sqrt{3} + \beta = 2\sqrt{3} \]
Expanding this:
\[ 2\sqrt{3} + 3\beta + \beta = 2\sqrt{3} \]
This simplifies to:
\[ 4\beta = 0 \implies \beta = 0 \]
Substituting \(\beta = 0\) into \(\alpha\):
\[ \alpha = 2 + \sqrt{3} \cdot 0 = 2 \]

#### Case 2:
\[ (2 + \sqrt{3} \beta) \sqrt{3} + \beta = -2\sqrt{3} \]
Expanding this:
\[ 2\sqrt{3} + 3\beta + \beta = -2\sqrt{3} \]
This simplifies to:
\[ 4\beta = -4\sqrt{3} \implies \beta = -\sqrt{3} \]
Substituting \(\beta = -\sqrt{3}\) into \(\alpha\):
\[ \alpha = 2 + \sqrt{3} \cdot (-\sqrt{3}) = 2 - 3 = -1 \]

### Step 8: Find the magnitudes of \(\alpha\)
Now we have two possible values for \(\alpha\):
1. \(|\alpha| = |2| = 2\)
2. \(|\alpha| = |-1| = 1\)

### Conclusion
The possible values of \(|\alpha|\) are \(2\) and \(1\).

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