Find image (reflection )of the point (7,4,-3) in the line `(x)/(1)=(y-1)/(2)=(z-2)/(3)`

Answer

To find the image (reflection) of the point \( P(7, 4, -3) \) in the line given by the equation

\[ \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}, \]

we will follow these steps:

### Step 1: Convert the line equation to parametric form
The line can be expressed in parametric form. Let \( \lambda \) be the parameter. Then, we can write:

\[ x = \lambda, \]
\[ y = 2\lambda + 1, \]
\[ z = 3\lambda + 2. \]

### Step 2: Find the coordinates of the point on the line
Let \( Q \) be a point on the line represented by the parameter \( \lambda \). Therefore, the coordinates of point \( Q \) are:

\[ Q(\lambda, 2\lambda + 1, 3\lambda + 2). \]

### Step 3: Find the direction ratios of line PQ
The direction ratios of the line segment \( PQ \) (from point \( P \) to point \( Q \)) are given by:

\[ PQ = Q - P = (\lambda - 7, (2\lambda + 1) - 4, (3\lambda + 2) + 3). \]

This simplifies to:

\[ PQ = (\lambda - 7, 2\lambda - 3, 3\lambda + 5). \]

### Step 4: Find the direction ratios of the line
The direction ratios of the line given by the equation are \( (1, 2, 3) \).

### Step 5: Set up the perpendicularity condition
Since \( PQ \) is perpendicular to the line, the dot product of the direction ratios must equal zero:

\[ (\lambda - 7) \cdot 1 + (2\lambda - 3) \cdot 2 + (3\lambda + 5) \cdot 3 = 0. \]

### Step 6: Solve the equation
Expanding the equation gives:

\[ \lambda - 7 + 4\lambda - 6 + 9\lambda + 15 = 0. \]

Combining like terms results in:

\[ 14\lambda + 2 = 0. \]

Solving for \( \lambda \):

\[ 14\lambda = -2 \implies \lambda = -\frac{1}{7}. \]

### Step 7: Find the coordinates of point Q
Substituting \( \lambda = -\frac{1}{7} \) back into the parametric equations for \( Q \):

\[ Q\left(-\frac{1}{7}, 2\left(-\frac{1}{7}\right) + 1, 3\left(-\frac{1}{7}\right) + 2\right). \]

Calculating the coordinates:

\[ Q\left(-\frac{1}{7}, -\frac{2}{7} + 1, -\frac{3}{7} + 2\right) = Q\left(-\frac{1}{7}, \frac{5}{7}, \frac{11}{7}\right). \]

### Step 8: Find the image point R
Since \( Q \) is the midpoint of \( P \) and \( R \), we can use the midpoint formula:

\[ Q = \left(\frac{7 + x}{2}, \frac{4 + y}{2}, \frac{-3 + z}{2}\right). \]

Setting this equal to the coordinates of \( Q \):

\[ \frac{7 + x}{2} = -\frac{1}{7}, \quad \frac{4 + y}{2} = \frac{5}{7}, \quad \frac{-3 + z}{2} = \frac{11}{7}. \]

### Step 9: Solve for x, y, z
1. For \( x \):

\[ 7 + x = -\frac{2}{7} \implies x = -\frac{2}{7} - 7 = -\frac{2}{7} - \frac{49}{7} = -\frac{51}{7}. \]

2. For \( y \):

\[ 4 + y = \frac{10}{7} \implies y = \frac{10}{7} - 4 = \frac{10}{7} - \frac{28}{7} = -\frac{18}{7}. \]

3. For \( z \):

\[ -3 + z = \frac{22}{7} \implies z = \frac{22}{7} + 3 = \frac{22}{7} + \frac{21}{7} = \frac{43}{7}. \]

### Final Result
Thus, the coordinates of the image point \( R \) are:

\[ R\left(-\frac{51}{7}, -\frac{18}{7}, \frac{43}{7}\right). \]