If three vectors `veca, vecb,vecc` are such that `veca ne 0` and `veca xx vecb = 2(veca xx vecc),|veca|=|vecc|=1, |vecb|=4` and the angle between `vecb` and `vecc` is `cos^(-1)(1//4)`, then `vecb-2vecc= lambda veca` where `lambda` is equal to:

Options

A. 4
B. `-4`
C. 2
D. `-2`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript while adding clarity and structure to each step.

### Step 1: Understand the given information
We have three vectors: \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The conditions provided are:
1. \(\vec{a} \neq 0\)
2. \(\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})\)
3. \(|\vec{a}| = |\vec{c}| = 1\)
4. \(|\vec{b}| = 4\)
5. The angle between \(\vec{b}\) and \(\vec{c}\) is \(\cos^{-1}(\frac{1}{4})\).

### Step 2: Rearranging the cross product equation
From the equation \(\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})\), we can rearrange this to:
\[ \vec{a} \times \vec{b} - 2(\vec{a} \times \vec{c}) = 0 \]
Factoring out \(\vec{a}\), we get:
\[ \vec{a} \times (\vec{b} - 2\vec{c}) = 0 \]
This implies that \(\vec{a}\) is parallel to \(\vec{b} - 2\vec{c}\). Therefore, we can write:
\[ \vec{b} - 2\vec{c} = \lambda \vec{a} \]
for some scalar \(\lambda\).

### Step 3: Squaring both sides
Now, we will square both sides of the equation:
\[ |\vec{b} - 2\vec{c}|^2 = |\lambda \vec{a}|^2 \]
This simplifies to:
\[ |\vec{b} - 2\vec{c}|^2 = \lambda^2 |\vec{a}|^2 \]
Since \(|\vec{a}| = 1\), we have:
\[ |\vec{b} - 2\vec{c}|^2 = \lambda^2 \]

### Step 4: Expanding the left-hand side
Now we expand the left-hand side:
\[ |\vec{b} - 2\vec{c}|^2 = |\vec{b}|^2 + |2\vec{c}|^2 - 2(\vec{b} \cdot 2\vec{c}) \]
This becomes:
\[ |\vec{b}|^2 + 4|\vec{c}|^2 - 4(\vec{b} \cdot \vec{c}) \]
Substituting the magnitudes:
\[ |\vec{b}|^2 = 4^2 = 16, \quad |\vec{c}|^2 = 1^2 = 1 \]
Thus:
\[ |\vec{b} - 2\vec{c}|^2 = 16 + 4 - 4(\vec{b} \cdot \vec{c}) \]
Now we need to find \(\vec{b} \cdot \vec{c}\).

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

### Step 6: Substitute back into the equation
Now substituting \(\vec{b} \cdot \vec{c} = 1\) into our equation:
\[ |\vec{b} - 2\vec{c}|^2 = 16 + 4 - 4 \cdot 1 = 16 + 4 - 4 = 16 \]
Thus:
\[ |\vec{b} - 2\vec{c}|^2 = 16 \]

### Step 7: Relating to \(\lambda\)
Now we have:
\[ \lambda^2 = 16 \]
Taking the square root gives:
\[ \lambda = \pm 4 \]

### Final Answer
Thus, the value of \(\lambda\) is \(4\) or \(-4\).

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