Let `veca` and `vecc` be unit vectors such that `|vecb|=4` and `vecaxxvecb=2(vecaxxvecc)`. The angle between `veca` and `vecc` is `cos^(-1)(1/4)`. If `vecb-2vecc=lamdaveca` then `lamda=`

Options

A. `1/3,1/4`
B. `-1/3,-1/4`
C. `3,-4`
D. `-3,4`

Explanation

To solve the problem, we will follow a step-by-step approach:

### Step 1: Understand the Given Information
We have:
- \(\vec{a}\) and \(\vec{c}\) are unit vectors, so \(|\vec{a}| = 1\) and \(|\vec{c}| = 1\).
- \(|\vec{b}| = 4\).
- The relationship given is \(\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})\).
- The angle between \(\vec{a}\) and \(\vec{c}\) is given by \(\cos^{-1}(1/4)\).

### Step 2: Express \(\vec{b}\) in terms of \(\vec{a}\) and \(\vec{c}\)
From the equation \(\vec{b} - 2\vec{c} = \lambda \vec{a}\), we can rearrange it to find \(\vec{b}\):
\[ \vec{b} = \lambda \vec{a} + 2\vec{c} \]

### Step 3: Calculate the Magnitude of \(\vec{b}\)
Taking the magnitude of both sides:
\[ |\vec{b}| = |\lambda \vec{a} + 2\vec{c}| \]
Since \(|\vec{b}| = 4\), we have:
\[ 4 = |\lambda \vec{a} + 2\vec{c}| \]

### Step 4: Use the Magnitude Formula
Using the formula for the magnitude of a vector:
\[ |\vec{b}|^2 = |\lambda \vec{a} + 2\vec{c}|^2 \]
Expanding this, we get:
\[ |\vec{b}|^2 = |\lambda \vec{a}|^2 + |2\vec{c}|^2 + 2(\lambda \vec{a} \cdot 2\vec{c}) \]
Substituting the magnitudes:
\[ 16 = \lambda^2 |\vec{a}|^2 + 4|\vec{c}|^2 + 4\lambda (\vec{a} \cdot \vec{c}) \]
Since \(|\vec{a}| = 1\) and \(|\vec{c}| = 1\), this simplifies to:
\[ 16 = \lambda^2 + 4 + 4\lambda (\vec{a} \cdot \vec{c}) \]

### Step 5: Find \(\vec{a} \cdot \vec{c}\)
Using the cosine of the angle between \(\vec{a}\) and \(\vec{c}\):
\[ \vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos(\theta) = 1 \cdot 1 \cdot \frac{1}{4} = \frac{1}{4} \]

### Step 6: Substitute Back into the Equation
Substituting \(\vec{a} \cdot \vec{c} = \frac{1}{4}\) into our equation:
\[ 16 = \lambda^2 + 4 + 4\lambda \left(\frac{1}{4}\right) \]
This simplifies to:
\[ 16 = \lambda^2 + 4 + \lambda \]
Rearranging gives:
\[ \lambda^2 + \lambda - 12 = 0 \]

### Step 7: Solve the Quadratic Equation
We can factor or use the quadratic formula to solve for \(\lambda\):
\[ \lambda^2 + \lambda - 12 = 0 \]
Factoring gives:
\[ (\lambda - 3)(\lambda + 4) = 0 \]
Thus, \(\lambda = 3\) or \(\lambda = -4\).

### Final Answer
The values of \(\lambda\) are \(3\) and \(-4\).