Consider the two vectors : `vecL=1hati+2hatj+3hatk " and " vecl=4hati+hatj+6hatk`. Find the value of the scalar `alpha ` such that the vector `vecL-alpha vecl` is perpendicular to `vecL`.

Answer

To solve the problem, we need to find the value of the scalar \( \alpha \) such that the vector \( \vec{L} - \alpha \vec{l} \) is perpendicular to \( \vec{L} \). We will use the property that two vectors are perpendicular if their dot product is zero.

### Step-by-Step Solution:

1. **Identify the vectors**:
- Given \( \vec{L} = 1\hat{i} + 2\hat{j} + 3\hat{k} \)
- Given \( \vec{l} = 4\hat{i} + 1\hat{j} + 6\hat{k} \)

2. **Form the vector \( \vec{L} - \alpha \vec{l} \)**:
\[ \vec{L} - \alpha \vec{l} = (1\hat{i} + 2\hat{j} + 3\hat{k}) - \alpha(4\hat{i} + 1\hat{j} + 6\hat{k}) \]
\[ = (1 - 4\alpha)\hat{i} + (2 - \alpha)\hat{j} + (3 - 6\alpha)\hat{k} \]

3. **Set up the dot product**:
We need to find \( \alpha \) such that:
\[ (\vec{L} - \alpha \vec{l}) \cdot \vec{L} = 0 \]
This means:
\[ ((1 - 4\alpha)\hat{i} + (2 - \alpha)\hat{j} + (3 - 6\alpha)\hat{k}) \cdot (1\hat{i} + 2\hat{j} + 3\hat{k}) = 0 \]

4. **Calculate the dot product**:
\[ = (1 - 4\alpha)(1) + (2 - \alpha)(2) + (3 - 6\alpha)(3) \]
\[ = (1 - 4\alpha) + 2(2 - \alpha) + 3(3 - 6\alpha) \]
\[ = 1 - 4\alpha + 4 - 2\alpha + 9 - 18\alpha \]
\[ = 1 + 4 + 9 - 4\alpha - 2\alpha - 18\alpha \]
\[ = 14 - 24\alpha \]

5. **Set the equation to zero**:
\[ 14 - 24\alpha = 0 \]
\[ 24\alpha = 14 \]
\[ \alpha = \frac{14}{24} = \frac{7}{12} \]

### Final Answer:
The value of the scalar \( \alpha \) is \( \frac{7}{12} \).