The coordinates of the foot of the perpendicular drawn from the point `A(1, 0,3 )` to the join of the points `B(4, 7, 1) and C(3, 5, 3)` are

Options

A. `((5)/(3), (7)/(3), (17)/(3))`
B. `((5, 7, 17))`
C. `((5)/(7), (-7)/(3), (17)/(3))`
D. `((-5)/(3), (7)/(3), (-17)/(3))`

Answer

To find the coordinates of the foot of the perpendicular drawn from the point \( A(1, 0, 3) \) to the line joining the points \( B(4, 7, 1) \) and \( C(3, 5, 3) \), we will follow these steps:

### Step 1: Find the direction ratios of the line BC
The direction ratios of the line joining points \( B(4, 7, 1) \) and \( C(3, 5, 3) \) can be calculated as follows:

\[ \text{Direction ratios} = (C_x - B_x, C_y - B_y, C_z - B_z) = (3 - 4, 5 - 7, 3 - 1) = (-1, -2, 2) \]

### Step 2: Write the parametric equations of the line BC
Using the point \( B(4, 7, 1) \) and the direction ratios, we can write the parametric equations of the line \( BC \):

\[ x = 4 - t, \quad y = 7 - 2t, \quad z = 1 + 2t \]

### Step 3: Find the coordinates of a general point on line BC
Let’s denote the coordinates of a general point \( D \) on the line \( BC \) as:

\[ D(t) = (4 - t, 7 - 2t, 1 + 2t) \]

### Step 4: Set up the perpendicularity condition
The vector \( AD \) from point \( A(1, 0, 3) \) to point \( D(t) \) is given by:

\[ AD = (D_x - A_x, D_y - A_y, D_z - A_z) = ((4 - t) - 1, (7 - 2t) - 0, (1 + 2t) - 3) = (3 - t, 7 - 2t, -2 + 2t) \]

The direction ratios of line \( BC \) are \( (-1, -2, 2) \). For \( AD \) to be perpendicular to \( BC \), their dot product must be zero:

\[ (3 - t)(-1) + (7 - 2t)(-2) + (-2 + 2t)(2) = 0 \]

### Step 5: Solve the dot product equation
Expanding the dot product gives:

\[ -(3 - t) - 2(7 - 2t) + 2(-2 + 2t) = 0 \]

This simplifies to:

\[ -t - 14 + 4t - 4 + 4t = 0 \]

Combining like terms:

\[ 7t - 18 = 0 \]

Thus,

\[ t = \frac{18}{7} \]

### Step 6: Substitute \( t \) back to find coordinates of point D
Now substituting \( t = \frac{18}{7} \) back into the parametric equations of line \( BC \):

\[ x = 4 - \frac{18}{7} = \frac{28 - 18}{7} = \frac{10}{7} \]
\[ y = 7 - 2 \cdot \frac{18}{7} = \frac{49 - 36}{7} = \frac{13}{7} \]
\[ z = 1 + 2 \cdot \frac{18}{7} = \frac{7 + 36}{7} = \frac{43}{7} \]

### Final Coordinates
Thus, the coordinates of the foot of the perpendicular from point \( A \) to line \( BC \) are:

\[ \left( \frac{10}{7}, \frac{13}{7}, \frac{43}{7} \right) \]