Let `alpha = (lambda-2) a + b` and `beta =(4lambda -2)a + 3b` be two given vectors where vectors a and b are non-collinear. The value of `lambda` for which vectors `alpha` and `beta` are collinear, is.

Options

A. 4
B. `-3`
C. 3
D. `-4`

Answer

To find the value of \( \lambda \) for which the vectors \( \alpha \) and \( \beta \) are collinear, we can follow these steps:

### Step 1: Write down the given vectors
The vectors are defined as:
\[ \alpha = (\lambda - 2) \mathbf{a} + \mathbf{b} \]
\[ \beta = (4\lambda - 2) \mathbf{a} + 3\mathbf{b} \]

### Step 2: Use the condition for collinearity
Two vectors \( \alpha \) and \( \beta \) are collinear if their cross product is zero:
\[ \alpha \times \beta = \mathbf{0} \]

### Step 3: Compute the cross product \( \alpha \times \beta \)
Substituting the expressions for \( \alpha \) and \( \beta \):
\[ \alpha \times \beta = \left((\lambda - 2) \mathbf{a} + \mathbf{b}\right) \times \left((4\lambda - 2) \mathbf{a} + 3\mathbf{b}\right) \]

### Step 4: Expand the cross product
Using the distributive property of the cross product:
\[ \alpha \times \beta = (\lambda - 2)(4\lambda - 2) \mathbf{a} \times \mathbf{a} + 3(\lambda - 2) \mathbf{a} \times \mathbf{b} + (4\lambda - 2) \mathbf{b} \times \mathbf{a} + 3 \mathbf{b} \times \mathbf{b} \]

### Step 5: Simplify using properties of the cross product
Recall that \( \mathbf{a} \times \mathbf{a} = \mathbf{0} \) and \( \mathbf{b} \times \mathbf{b} = \mathbf{0} \):
\[ \alpha \times \beta = 3(\lambda - 2) \mathbf{a} \times \mathbf{b} + (4\lambda - 2) \mathbf{b} \times \mathbf{a} \]
Using the property \( \mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b} \):
\[ \alpha \times \beta = 3(\lambda - 2) \mathbf{a} \times \mathbf{b} - (4\lambda - 2) \mathbf{a} \times \mathbf{b} \]

### Step 6: Factor out \( \mathbf{a} \times \mathbf{b} \)
\[ \alpha \times \beta = \left(3(\lambda - 2) - (4\lambda - 2)\right) \mathbf{a} \times \mathbf{b} \]

### Step 7: Set the expression to zero
Since \( \mathbf{a} \times \mathbf{b} \neq \mathbf{0} \) (because \( \mathbf{a} \) and \( \mathbf{b} \) are non-collinear), we can set the scalar part to zero:
\[ 3(\lambda - 2) - (4\lambda - 2) = 0 \]

### Step 8: Solve for \( \lambda \)
Expanding and simplifying:
\[ 3\lambda - 6 - 4\lambda + 2 = 0 \]
\[ - \lambda - 4 = 0 \]
\[ \lambda = -4 \]

### Final Answer
The value of \( \lambda \) for which the vectors \( \alpha \) and \( \beta \) are collinear is:
\[ \lambda = -4 \]

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