`veca and vecc` are unit vectors and `|vecb|=4` the angle between `veca and vecb is cos ^(-1)(1//4) and vecb - 2vecc=lambdaveca` the value of `lambda` is

Options

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

Explanation

To solve the problem step by step, we start with the given information and derive the value of \( \lambda \).

### Step 1: Understand the given information
We know:
- \( \vec{a} \) and \( \vec{c} \) are unit vectors, meaning \( |\vec{a}| = 1 \) and \( |\vec{c}| = 1 \).
- The magnitude of \( \vec{b} \) is \( |\vec{b}| = 4 \).
- The angle between \( \vec{a} \) and \( \vec{b} \) is given by \( \cos^{-1}\left(\frac{1}{4}\right) \).
- The equation \( \vec{b} - 2\vec{c} = \lambda \vec{a} \).

### Step 2: Rearranging the equation
From the equation \( \vec{b} - 2\vec{c} = \lambda \vec{a} \), we can express \( \vec{b} \) as:
\[ \vec{b} = 2\vec{c} + \lambda \vec{a} \]

### Step 3: Taking the modulus and squaring both sides
Taking the modulus of both sides gives:
\[ |\vec{b}|^2 = |2\vec{c} + \lambda \vec{a}|^2 \]
Since \( |\vec{b}| = 4 \), we have:
\[ 16 = |2\vec{c} + \lambda \vec{a}|^2 \]

### Step 4: Expanding the right-hand side
Using the property of dot products, we can expand the right-hand side:
\[ |2\vec{c} + \lambda \vec{a}|^2 = |2\vec{c}|^2 + |\lambda \vec{a}|^2 + 2(2\vec{c} \cdot \lambda \vec{a}) \]
Calculating each term:
- \( |2\vec{c}|^2 = 4|\vec{c}|^2 = 4 \cdot 1 = 4 \)
- \( |\lambda \vec{a}|^2 = \lambda^2 |\vec{a}|^2 = \lambda^2 \cdot 1 = \lambda^2 \)
- The dot product \( 2\vec{c} \cdot \lambda \vec{a} = 2\lambda (\vec{c} \cdot \vec{a}) \)

### Step 5: Substitute the dot product
We know from the problem that the angle between \( \vec{a} \) and \( \vec{c} \) gives us \( \cos(\theta) = \frac{1}{4} \). Therefore, we have:
\[ \vec{c} \cdot \vec{a} = \frac{1}{4} \]
Substituting this into our equation gives:
\[ 16 = 4 + \lambda^2 + 2\lambda \cdot 2 \cdot \frac{1}{4} \]
This simplifies to:
\[ 16 = 4 + \lambda^2 + \lambda \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ \lambda^2 + \lambda + 4 - 16 = 0 \]
\[ \lambda^2 + \lambda - 12 = 0 \]

### Step 7: Solving the quadratic equation
Now we can solve the quadratic equation \( \lambda^2 + \lambda - 12 = 0 \) using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = -12 \):
\[ \lambda = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \]
\[ \lambda = \frac{-1 \pm \sqrt{1 + 48}}{2} \]
\[ \lambda = \frac{-1 \pm \sqrt{49}}{2} \]
\[ \lambda = \frac{-1 \pm 7}{2} \]

### Step 8: Finding the values of \( \lambda \)
Calculating the two possible values:
1. \( \lambda = \frac{6}{2} = 3 \)
2. \( \lambda = \frac{-8}{2} = -4 \)

Thus, the values of \( \lambda \) are \( 3 \) and \( -4 \).

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

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