If `vec(a)=3hat(i)+hat(j)+9hat(k)` and `vec(b)=hat(i)+lambda hat(j)+3hat(k)`, then find the value of `'lambda'` for which the vectors `(vec(a)+vec(b))` and `(vec(a)-vec(b))` are perpendicular to each other.

Answer

To find the value of \( \lambda \) for which the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) are perpendicular, we follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{a} = 3\hat{i} + \hat{j} + 9\hat{k} \]
\[ \vec{b} = \hat{i} + \lambda \hat{j} + 3\hat{k} \]

### Step 2: Calculate \( \vec{a} + \vec{b} \)
\[ \vec{a} + \vec{b} = (3\hat{i} + \hat{j} + 9\hat{k}) + (\hat{i} + \lambda \hat{j} + 3\hat{k}) \]
Combine the components:
\[ = (3 + 1)\hat{i} + (1 + \lambda)\hat{j} + (9 + 3)\hat{k} \]
\[ = 4\hat{i} + (1 + \lambda)\hat{j} + 12\hat{k} \]

### Step 3: Calculate \( \vec{a} - \vec{b} \)
\[ \vec{a} - \vec{b} = (3\hat{i} + \hat{j} + 9\hat{k}) - (\hat{i} + \lambda \hat{j} + 3\hat{k}) \]
Combine the components:
\[ = (3 - 1)\hat{i} + (1 - \lambda)\hat{j} + (9 - 3)\hat{k} \]
\[ = 2\hat{i} + (1 - \lambda)\hat{j} + 6\hat{k} \]

### Step 4: Set up the dot product
For the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \) to be perpendicular, their dot product must equal zero:
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0 \]

Calculating the dot product:
\[ (4\hat{i} + (1 + \lambda)\hat{j} + 12\hat{k}) \cdot (2\hat{i} + (1 - \lambda)\hat{j} + 6\hat{k}) \]

### Step 5: Compute the dot product
\[ = 4 \cdot 2 + (1 + \lambda)(1 - \lambda) + 12 \cdot 6 \]
Calculating each term:
\[ = 8 + (1 - \lambda^2) + 72 \]
Combine the terms:
\[ = 8 + 72 + 1 - \lambda^2 = 81 - \lambda^2 \]

### Step 6: Set the equation to zero
Setting the dot product to zero:
\[ 81 - \lambda^2 = 0 \]

### Step 7: Solve for \( \lambda \)
\[ \lambda^2 = 81 \]
Taking the square root:
\[ \lambda = \pm 9 \]

### Final Answer
The values of \( \lambda \) are:
\[ \lambda = 9 \quad \text{or} \quad \lambda = -9 \]

---