The value of `lamda` so that the unit vectors `(2hat(i) + lamda hat(j) + hat(k))/(sqrt(5+ lamda^(2))) and (hat(i) - 2hat(j) + 3hat(k))/(sqrt14)` are orthogonal is

Explanation

To find the value of `lambda` such that the given unit vectors are orthogonal, we will follow these steps:

### Step 1: Define the Unit Vectors
The two unit vectors given are:
1. \( \mathbf{A} = \frac{2\hat{i} + \lambda \hat{j} + \hat{k}}{\sqrt{5 + \lambda^2}} \)
2. \( \mathbf{B} = \frac{\hat{i} - 2\hat{j} + 3\hat{k}}{\sqrt{14}} \)

### Step 2: Use the Orthogonality Condition
Two vectors are orthogonal if their dot product is zero. Therefore, we need to compute the dot product \( \mathbf{A} \cdot \mathbf{B} \) and set it to zero:
\[ \mathbf{A} \cdot \mathbf{B} = 0 \]

### Step 3: Calculate the Dot Product
The dot product is given by:
\[ \mathbf{A} \cdot \mathbf{B} = \left(\frac{2\hat{i} + \lambda \hat{j} + \hat{k}}{\sqrt{5 + \lambda^2}}\right) \cdot \left(\frac{\hat{i} - 2\hat{j} + 3\hat{k}}{\sqrt{14}}\right) \]
Calculating the dot product:
\[ \mathbf{A} \cdot \mathbf{B} = \frac{1}{\sqrt{5 + \lambda^2} \cdot \sqrt{14}} \left( 2 \cdot 1 + \lambda \cdot (-2) + 1 \cdot 3 \right) \]
This simplifies to:
\[ \mathbf{A} \cdot \mathbf{B} = \frac{1}{\sqrt{5 + \lambda^2} \cdot \sqrt{14}} (2 - 2\lambda + 3) = \frac{5 - 2\lambda}{\sqrt{5 + \lambda^2} \cdot \sqrt{14}} \]

### Step 4: Set the Dot Product to Zero
For the vectors to be orthogonal:
\[ \frac{5 - 2\lambda}{\sqrt{5 + \lambda^2} \cdot \sqrt{14}} = 0 \]
This implies:
\[ 5 - 2\lambda = 0 \]

### Step 5: Solve for `lambda`
Rearranging gives:
\[ 2\lambda = 5 \implies \lambda = \frac{5}{2} = 2.5 \]

### Conclusion
The value of `lambda` so that the unit vectors are orthogonal is:
\[ \lambda = 2.5 \]

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