If the foot of perpendicular drawn from the point (2, 5, 1) on a line passing through `(alpha, 2alpha, 5)" is "((1)/(5),(2)/(5),(3)/(5))`, then `alpha` is equal to

Options

A. `(19)/(9)`
B. `(11)/(570)`
C. `(57)/(54)`
D. `(1)/(25)`

Answer

To find the value of \( \alpha \) given the conditions of the problem, we can follow these steps:

### Step 1: Identify the Points
We have the following points:
- Point A (from which the perpendicular is drawn): \( (2, 5, 1) \)
- Foot of the perpendicular (Point B): \( \left( \frac{1}{5}, \frac{2}{5}, \frac{3}{5} \right) \)
- Point C (on the line): \( (\alpha, 2\alpha, 5) \)

### Step 2: Calculate Direction Ratios
To find the direction ratios of the lines AB and BC, we first calculate the vectors.

**Vector AB:**
\[ AB = B - A = \left( \frac{1}{5} - 2, \frac{2}{5} - 5, \frac{3}{5} - 1 \right) = \left( \frac{1}{5} - \frac{10}{5}, \frac{2}{5} - \frac{25}{5}, \frac{3}{5} - \frac{5}{5} \right) = \left( -\frac{9}{5}, -\frac{23}{5}, -\frac{2}{5} \right) \]

**Vector BC:**
\[ BC = C - B = \left( \alpha - \frac{1}{5}, 2\alpha - \frac{2}{5}, 5 - \frac{3}{5} \right) = \left( \alpha - \frac{1}{5}, 2\alpha - \frac{2}{5}, \frac{22}{5} \right) \]

### Step 3: Use the Perpendicular Condition
Since AB and BC are perpendicular, their dot product must equal zero:
\[ AB \cdot BC = 0 \]

Calculating the dot product:
\[ \left( -\frac{9}{5}, -\frac{23}{5}, -\frac{2}{5} \right) \cdot \left( \alpha - \frac{1}{5}, 2\alpha - \frac{2}{5}, \frac{22}{5} \right) = 0 \]

This expands to:
\[ -\frac{9}{5} \left( \alpha - \frac{1}{5} \right) - \frac{23}{5} \left( 2\alpha - \frac{2}{5} \right) - \frac{2}{5} \left( \frac{22}{5} \right) = 0 \]

### Step 4: Simplify the Equation
Expanding the terms:
\[ -\frac{9\alpha}{5} + \frac{9}{25} - \frac{46\alpha}{5} + \frac{46}{25} - \frac{44}{25} = 0 \]

Combining like terms:
\[ -\frac{55\alpha}{5} + \frac{9 + 46 - 44}{25} = 0 \]
\[ -\frac{55\alpha}{5} + \frac{11}{25} = 0 \]

### Step 5: Solve for \( \alpha \)
Multiplying through by 25 to eliminate the fraction:
\[ -275\alpha + 11 = 0 \]
\[ 275\alpha = 11 \]
\[ \alpha = \frac{11}{275} = \frac{1}{25} \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = \frac{1}{25} \]

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