Let a, b and c be three vectors satisfying `a xx b=(a xx c), |a| =|c| =1, |b|=4 and |b xx c| = sqrt(15)`. If `b-2c=lambda a`, then `lambda` equals

Options

A. 1
B. `-1`
C. 2
D. `+-4`

Answer

To solve the problem, we need to find the value of \(\lambda\) given the vectors \(a\), \(b\), and \(c\) with the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know that \( |a| = |c| = 1 \) (both vectors \(a\) and \(c\) are unit vectors).
- The magnitude of vector \(b\) is \( |b| = 4 \).
- The magnitude of the cross product \( |b \times c| = \sqrt{15} \).
- We also have the equation \( b - 2c = \lambda a \).

2. **Finding the Magnitude of \(b - 2c\)**:
- We can square both sides of the equation \( b - 2c = \lambda a \):
\[ |b - 2c|^2 = |\lambda a|^2 \]
- Since \( |a| = 1 \), we have:
\[ |b - 2c|^2 = \lambda^2 \]

3. **Expanding the Left Side**:
- We can expand \( |b - 2c|^2 \):
\[ |b - 2c|^2 = |b|^2 + |2c|^2 - 2(b \cdot 2c) \]
- Substituting the known magnitudes:
\[ |b|^2 = 4^2 = 16, \quad |2c|^2 = 2^2 |c|^2 = 4 \cdot 1 = 4 \]
- Thus:
\[ |b - 2c|^2 = 16 + 4 - 4(b \cdot c) \]

4. **Finding \(b \cdot c\)**:
- We know that \( |b \times c| = |b||c|\sin\theta \) where \(\theta\) is the angle between \(b\) and \(c\):
\[ |b \times c| = 4 \cdot 1 \cdot \sin\theta = 4\sin\theta \]
- Given \( |b \times c| = \sqrt{15} \), we have:
\[ 4\sin\theta = \sqrt{15} \implies \sin\theta = \frac{\sqrt{15}}{4} \]
- Using the identity \( \cos^2\theta + \sin^2\theta = 1 \):
\[ \cos^2\theta = 1 - \left(\frac{\sqrt{15}}{4}\right)^2 = 1 - \frac{15}{16} = \frac{1}{16} \implies \cos\theta = \frac{1}{4} \]

5. **Substituting Back**:
- Now substituting \( b \cdot c = |b||c|\cos\theta = 4 \cdot 1 \cdot \frac{1}{4} = 1 \):
\[ |b - 2c|^2 = 16 + 4 - 4 \cdot 1 = 16 + 4 - 4 = 16 \]

6. **Equating to Find \(\lambda\)**:
- Now we have:
\[ \lambda^2 = 16 \implies \lambda = \pm 4 \]

### Final Answer:
\[ \lambda = 4 \text{ or } \lambda = -4 \]