The magnitude of the vectors `vec(a) and vec(b)` are equal and the angle between them is `60^(@)`. If the vectors `lambda vec(a)+vec(b) and vec(a)-lambda vec(b)` are perpendicular to each other, then what is the value of `lambda`?

Options

A. 1
B. 2
C. 3
D. 4

Explanation

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

### Step 1: Understand the Given Information
We know:
1. The magnitudes of vectors \( \vec{a} \) and \( \vec{b} \) are equal, i.e., \( |\vec{a}| = |\vec{b}| \).
2. The angle between them is \( 60^\circ \).
3. The vectors \( \lambda \vec{a} + \vec{b} \) and \( \vec{a} - \lambda \vec{b} \) are perpendicular.

### Step 2: Use the Condition of Perpendicularity
For two vectors to be perpendicular, their dot product must equal zero:
\[ (\lambda \vec{a} + \vec{b}) \cdot (\vec{a} - \lambda \vec{b}) = 0 \]

### Step 3: Expand the Dot Product
Expanding the dot product, we have:
\[ \lambda \vec{a} \cdot \vec{a} - \lambda^2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \lambda \vec{b} \cdot \vec{b} = 0 \]
This simplifies to:
\[ \lambda |\vec{a}|^2 - \lambda^2 (\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{a} - \lambda |\vec{b}|^2 = 0 \]

### Step 4: Substitute Known Values
Since \( |\vec{a}| = |\vec{b}| \), we can denote this common magnitude as \( k \):
\[ |\vec{a}|^2 = k^2 \quad \text{and} \quad |\vec{b}|^2 = k^2 \]
Also, using the cosine of the angle between them:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(60^\circ) = k^2 \cdot \frac{1}{2} = \frac{k^2}{2} \]

### Step 5: Substitute into the Dot Product Equation
Substituting these values into our equation gives:
\[ \lambda k^2 - \lambda^2 \left(\frac{k^2}{2}\right) + \frac{k^2}{2} - \lambda k^2 = 0 \]
This simplifies to:
\[ -\frac{\lambda^2 k^2}{2} + \frac{k^2}{2} = 0 \]

### Step 6: Factor Out \( k^2 \)
Since \( k^2 \neq 0 \), we can factor it out:
\[ -\frac{\lambda^2}{2} + \frac{1}{2} = 0 \]
Multiplying through by \( 2 \) gives:
\[ -\lambda^2 + 1 = 0 \]

### Step 7: Solve for \( \lambda \)
Rearranging gives:
\[ \lambda^2 = 1 \]
Taking the square root yields:
\[ \lambda = \pm 1 \]

### Conclusion
Thus, the possible values for \( \lambda \) are \( 1 \) and \( -1 \). However, since the problem may require a positive value, we conclude:
\[ \lambda = 1 \]