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

Options

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

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the vectors \( \lambda \vec{a} + \vec{b} \) and \( \vec{a} - \lambda \vec{b} \) are perpendicular to each other. Given that the magnitudes of the vectors \( \vec{a} \) and \( \vec{b} \) are equal and the angle between them is \( 60^\circ \), we can follow these steps:

### Step 1: Set up the dot product condition for perpendicular vectors
Since the vectors \( \lambda \vec{a} + \vec{b} \) and \( \vec{a} - \lambda \vec{b} \) are perpendicular, their dot product must equal zero:
\[ (\lambda \vec{a} + \vec{b}) \cdot (\vec{a} - \lambda \vec{b}) = 0 \]

### Step 2: Expand the dot product
Using the distributive property of the dot product, we can expand the expression:
\[ \lambda \vec{a} \cdot \vec{a} - \lambda^2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \lambda \vec{b} \cdot \vec{b} = 0 \]
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 3: Substitute known values
Since \( |\vec{a}| = |\vec{b}| \), we denote this common magnitude as \( k \). Thus, \( |\vec{a}|^2 = |\vec{b}|^2 = k^2 \). Also, since the angle between \( \vec{a} \) and \( \vec{b} \) is \( 60^\circ \), we can express \( \vec{a} \cdot \vec{b} \) as:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(60^\circ) = k^2 \cdot \frac{1}{2} = \frac{k^2}{2} \]

### Step 4: 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 5: Factor out \( k^2/2 \)
Factoring out \( \frac{k^2}{2} \) from the equation:
\[ \frac{k^2}{2} (1 - \lambda^2) = 0 \]
Since \( k^2 \) is not zero (as it represents the magnitude of the vectors), we have:
\[ 1 - \lambda^2 = 0 \]

### Step 6: Solve for \( \lambda \)
This leads us to:
\[ \lambda^2 = 1 \]
Taking the square root gives:
\[ \lambda = \pm 1 \]

### Conclusion
Since the options provided include only positive values, we conclude:
\[ \lambda = 1 \]