Let A(1, 1, 1), B(3, 7, 4) and C (-1, 3, 0) be three
points in a plane.
The co-ordinates of the foot of perpendicular drawn
from point C to the line segment AB is

Options

A. `(59/49, 79/49, 64/(49))`
B. `((-59)/49, 79/49, 64/49)`
C. `(59/49, (_79)/49, 64/49)`
D. `(59/49, 79/49, (-64)/49)`

Answer

To find the coordinates of the foot of the perpendicular drawn from point C to the line segment AB, we can follow these steps:

### Step 1: Identify the Points
We have the following points:
- A(1, 1, 1)
- B(3, 7, 4)
- C(-1, 3, 0)

### Step 2: Find the Direction Ratios of Line Segment AB
The direction ratios of line segment AB can be calculated as follows:
- Direction ratios = B - A = (3 - 1, 7 - 1, 4 - 1) = (2, 6, 3)

### Step 3: Write the Parametric Equations of Line AB
Using point A and the direction ratios, we can write the parametric equations for the line segment AB:
- x = 1 + 2λ
- y = 1 + 6λ
- z = 1 + 3λ

### Step 4: Define Point P on Line AB
Let P be any point on line segment AB. Thus, we can express the coordinates of point P in terms of λ:
- P(1 + 2λ, 1 + 6λ, 1 + 3λ)

### Step 5: Find the Direction Ratios of CP
The direction ratios of line segment CP can be calculated as:
- CP = P - C = ((1 + 2λ) - (-1), (1 + 6λ) - 3, (1 + 3λ) - 0)
- This simplifies to: (2λ + 2, 6λ - 2, 3λ + 1)

### Step 6: Set Up the Perpendicular Condition
Since CP is perpendicular to AB, the dot product of the direction ratios of AB and CP must equal zero:
- (2, 6, 3) • (2λ + 2, 6λ - 2, 3λ + 1) = 0
- This gives us the equation:
\( 2(2λ + 2) + 6(6λ - 2) + 3(3λ + 1) = 0 \)

### Step 7: Expand and Simplify the Equation
Expanding the equation:
- \( 4λ + 4 + 36λ - 12 + 9λ + 3 = 0 \)
- Combine like terms:
\( 49λ - 5 = 0 \)

### Step 8: Solve for λ
Solving for λ:
- \( 49λ = 5 \)
- \( λ = \frac{5}{49} \)

### Step 9: Substitute λ Back to Find Point P
Substituting λ back into the parametric equations for P:
- \( x = 1 + 2\left(\frac{5}{49}\right) = 1 + \frac{10}{49} = \frac{49 + 10}{49} = \frac{59}{49} \)
- \( y = 1 + 6\left(\frac{5}{49}\right) = 1 + \frac{30}{49} = \frac{49 + 30}{49} = \frac{79}{49} \)
- \( z = 1 + 3\left(\frac{5}{49}\right) = 1 + \frac{15}{49} = \frac{49 + 15}{49} = \frac{64}{49} \)

### Final Coordinates of Point P
Thus, the coordinates of the foot of the perpendicular from point C to line segment AB are:
- P\(\left(\frac{59}{49}, \frac{79}{49}, \frac{64}{49}\right)\)

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