`veca and vecc` are unit vectors and `|vecb|=4` the angle between `veca and vecc` `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 will follow the instructions provided in the video transcript.

### 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 angle \(\theta\) between \(\vec{a}\) and \(\vec{c}\) 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 magnitude of both sides
Now, we take the magnitude of both sides:
\[ |\vec{b}| = |2\vec{c} + \lambda \vec{a}| \]
Given that \(|\vec{b}| = 4\), we have:
\[ 4 = |2\vec{c} + \lambda \vec{a}| \]

### Step 4: Squaring both sides
Squaring both sides gives:
\[ 16 = |2\vec{c} + \lambda \vec{a}|^2 \]
Using the property of magnitudes, we can expand the right side:
\[ |2\vec{c} + \lambda \vec{a}|^2 = |2\vec{c}|^2 + |\lambda \vec{a}|^2 + 2(2\vec{c} \cdot \lambda \vec{a}) \]
This simplifies to:
\[ |2\vec{c}|^2 = 4|\vec{c}|^2 = 4 \quad \text{(since } |\vec{c}| = 1\text{)} \]
\[ |\lambda \vec{a}|^2 = \lambda^2 |\vec{a}|^2 = \lambda^2 \quad \text{(since } |\vec{a}| = 1\text{)} \]
Thus, we have:
\[ 16 = 4 + \lambda^2 + 4\lambda(\vec{c} \cdot \vec{a}) \]

### Step 5: Substitute \(\vec{c} \cdot \vec{a}\)
Since \(\cos \theta = \frac{1}{4}\), we know:
\[ \vec{c} \cdot \vec{a} = |\vec{c}||\vec{a}|\cos \theta = 1 \cdot 1 \cdot \frac{1}{4} = \frac{1}{4} \]
Substituting this into the equation gives:
\[ 16 = 4 + \lambda^2 + 4\lambda \left(\frac{1}{4}\right) \]
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 factor the quadratic equation:
\[ \lambda^2 + 4\lambda - 3\lambda - 12 = 0 \]
Factoring gives:
\[ (\lambda + 4)(\lambda - 3) = 0 \]
Thus, the solutions for \(\lambda\) are:
\[ \lambda = -4 \quad \text{or} \quad \lambda = 3 \]

### Step 8: Conclusion
The value of \(\lambda\) is either \(3\) or \(-4\). Since the problem states that we need the value of \(\lambda\), we conclude:
\[ \lambda = -4 \]