The image (or reflection) of the point (1,2-1) in the plane `vecr.(3hati-5hatj+4hatk)=5` is

Options

A. `(73//5,-6//4,39/25)`
B. `(73//25,6//5,39//25)`
C. `(-1,-2,1)`
D. none of these

Explanation

To find the image of the point \( P(1, 2, -1) \) in the plane given by the equation \( \vec{r} \cdot (3\hat{i} - 5\hat{j} + 4\hat{k}) = 5 \), we will follow these steps:

### Step 1: Identify the normal vector of the plane
The normal vector \( \vec{n} \) to the plane can be directly obtained from the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) in the plane equation:
\[ \vec{n} = 3\hat{i} - 5\hat{j} + 4\hat{k} \]

### Step 2: Find the foot of the perpendicular (point Q)
To find the foot of the perpendicular from point \( P \) to the plane, we need the equation of the line passing through \( P \) in the direction of the normal vector \( \vec{n} \).

Using the parametric equations of the line:
\[ \begin{align*} x &= 1 + 3\lambda \\ y &= 2 - 5\lambda \\ z &= -1 + 4\lambda \end{align*} \]

### Step 3: Substitute into the plane equation
Substituting the parametric equations into the plane equation:
\[ 3(1 + 3\lambda) - 5(2 - 5\lambda) + 4(-1 + 4\lambda) = 5 \]
Expanding this gives:
\[ 3 + 9\lambda - 10 + 25\lambda - 4 + 16\lambda = 5 \]
Combining like terms:
\[ (9\lambda + 25\lambda + 16\lambda) + (3 - 10 - 4) = 5 \]
\[ 50\lambda - 11 = 5 \]
\[ 50\lambda = 16 \implies \lambda = \frac{16}{50} = \frac{8}{25} \]

### Step 4: Find the coordinates of point Q
Substituting \( \lambda = \frac{8}{25} \) back into the parametric equations:
\[ \begin{align*} x_Q &= 1 + 3\left(\frac{8}{25}\right) = 1 + \frac{24}{25} = \frac{49}{25} \\ y_Q &= 2 - 5\left(\frac{8}{25}\right) = 2 - \frac{40}{25} = \frac{10}{5} - \frac{40}{25} = \frac{2}{5} \\ z_Q &= -1 + 4\left(\frac{8}{25}\right) = -1 + \frac{32}{25} = -\frac{25}{25} + \frac{32}{25} = \frac{7}{25} \end{align*} \]
Thus, the coordinates of point \( Q \) are \( Q\left(\frac{49}{25}, \frac{2}{5}, \frac{7}{25}\right) \).

### Step 5: Find the coordinates of the image point R
Since \( Q \) is the midpoint of \( P \) and \( R \), we can use the midpoint formula:
\[ Q = \left(\frac{x_P + x_R}{2}, \frac{y_P + y_R}{2}, \frac{z_P + z_R}{2}\right) \]
Setting up the equations:
\[ \frac{49}{25} = \frac{1 + x_R}{2} \implies 49 = 25 + 25x_R \implies 25x_R = 24 \implies x_R = \frac{24}{25} \]
\[ \frac{2}{5} = \frac{2 + y_R}{2} \implies 2 = 2 + y_R \implies y_R = -\frac{6}{5} \]
\[ \frac{7}{25} = \frac{-1 + z_R}{2} \implies 7 = -25 + 25z_R \implies 25z_R = 32 \implies z_R = \frac{32}{25} \]

### Final Result
The image of the point \( P(1, 2, -1) \) in the plane is:
\[ R\left(\frac{73}{25}, -\frac{6}{5}, \frac{39}{25}\right) \]