Let `veca,vecb,vecc` be three vectors such that `veca ne vec0` and `veca xx vecb=2veca xx vecc, |veca|=|vecc|=1,|vecb|=4 and |vecbxxvecc|=sqrt(15)`. If `vecb-2vecc=lambdaveca`, then `|lambda|` is equal to :

Explanation

To solve the problem step by step, let's break it down:

### Given:
1. \(\vec{a}, \vec{b}, \vec{c}\) are vectors such that \(\vec{a} \neq \vec{0}\).
2. \(\vec{a} \times \vec{b} = 2 \vec{a} \times \vec{c}\)
3. \(|\vec{a}| = |\vec{c}| = 1\)
4. \(|\vec{b}| = 4\)
5. \(|\vec{b} \times \vec{c}| = \sqrt{15}\)
6. \(\vec{b} - 2\vec{c} = \lambda \vec{a}\)

### Objective:
Find \(|\lambda|\).

### Step 1: Analyze the cross product condition
From the equation \(\vec{a} \times \vec{b} = 2 \vec{a} \times \vec{c}\), we can take the magnitude of both sides:
\[ |\vec{a} \times \vec{b}| = 2 |\vec{a} \times \vec{c}| \]
Using the property of the cross product, we know:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta_{ab} \]
\[ |\vec{a} \times \vec{c}| = |\vec{a}| |\vec{c}| \sin \theta_{ac} \]
where \(\theta_{ab}\) is the angle between \(\vec{a}\) and \(\vec{b}\), and \(\theta_{ac}\) is the angle between \(\vec{a}\) and \(\vec{c}\).

Since \(|\vec{a}| = |\vec{c}| = 1\):
\[ |\vec{a} \times \vec{b}| = 4 \sin \theta_{ab} \]
\[ |\vec{a} \times \vec{c}| = \sin \theta_{ac} \]
Thus, we have:
\[ 4 \sin \theta_{ab} = 2 \sin \theta_{ac} \]
This simplifies to:
\[ 2 \sin \theta_{ab} = \sin \theta_{ac} \]

### Step 2: Use the given magnitude of the cross product
We know:
\[ |\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin \theta_{bc} \]
Substituting the known magnitudes:
\[ \sqrt{15} = 4 \cdot 1 \cdot \sin \theta_{bc} \]
This gives:
\[ \sin \theta_{bc} = \frac{\sqrt{15}}{4} \]

### Step 3: Find \(\cos \theta_{bc}\)
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \sin^2 \theta_{bc} = \left(\frac{\sqrt{15}}{4}\right)^2 = \frac{15}{16} \]
Thus,
\[ \cos^2 \theta_{bc} = 1 - \frac{15}{16} = \frac{1}{16} \]
So,
\[ \cos \theta_{bc} = \frac{1}{4} \]

### Step 4: Substitute into the equation for \(\lambda\)
From the equation \(\vec{b} - 2\vec{c} = \lambda \vec{a}\), we can square both sides:
\[ |\vec{b} - 2\vec{c}|^2 = |\lambda \vec{a}|^2 \]
This gives:
\[ |\vec{b}|^2 - 4 \vec{b} \cdot \vec{c} + 4 |\vec{c}|^2 = \lambda^2 \]
Substituting the known values:
\[ 16 - 4 \vec{b} \cdot \vec{c} + 4 = \lambda^2 \]
\[ 20 - 4 \vec{b} \cdot \vec{c} = \lambda^2 \]

### Step 5: Find \(\vec{b} \cdot \vec{c}\)
Using the dot product:
\[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta_{bc} = 4 \cdot 1 \cdot \frac{1}{4} = 1 \]
Substituting back:
\[ 20 - 4 \cdot 1 = \lambda^2 \]
\[ 20 - 4 = \lambda^2 \]
\[ \lambda^2 = 16 \]
Thus,
\[ |\lambda| = 4 \]

### Final Answer:
\(|\lambda| = 4\)