Image of line `(x - 2)/3 = (y - 1)/1 = (z - 1)/(-4)` in the plane `x + y + z = 7` is

Options

A. `(x - 4)/1 = (y - 3)/(1) = (z- 3)/(1)`
B. `(x - 3)/1 = (y - 4)/(1) = (z- 3)/(1)`
C. `(x - 4)/3 = (y - 3)/(1) = (z- 3)/(-4)`
D. `(x - 3)/1 = (y - 4)/(1) = (z- 3)/(-4)`

Answer

To find the image of the line given by the equation \((x - 2)/3 = (y - 1)/1 = (z - 1)/(-4)\) in the plane defined by \(x + y + z = 7\), we can follow these steps:

### Step 1: Identify the direction ratios of the line and the normal to the plane.
The line can be represented in parametric form as:
- \(x = 2 + 3t\)
- \(y = 1 + t\)
- \(z = 1 - 4t\)

From this, we can identify the direction ratios of the line as \(d_1 = 3\), \(d_2 = 1\), and \(d_3 = -4\).

The normal vector to the plane \(x + y + z = 7\) is given by the coefficients of \(x\), \(y\), and \(z\), which are \(1, 1, 1\). Thus, the direction ratios of the normal to the plane are \(n_1 = 1\), \(n_2 = 1\), and \(n_3 = 1\).

### Step 2: Check the angle between the line and the plane.
To find the angle between the line and the plane, we can use the dot product of the direction ratios of the line and the normal to the plane.

The cosine of the angle \(\theta\) between the line and the normal is given by:
\[ \cos \theta = \frac{d_1 \cdot n_1 + d_2 \cdot n_2 + d_3 \cdot n_3}{\sqrt{d_1^2 + d_2^2 + d_3^2} \cdot \sqrt{n_1^2 + n_2^2 + n_3^2}} \]
Substituting the values:
\[ \cos \theta = \frac{3 \cdot 1 + 1 \cdot 1 + (-4) \cdot 1}{\sqrt{3^2 + 1^2 + (-4)^2} \cdot \sqrt{1^2 + 1^2 + 1^2}} = \frac{3 + 1 - 4}{\sqrt{9 + 1 + 16} \cdot \sqrt{3}} = \frac{0}{\sqrt{26} \cdot \sqrt{3}} = 0 \]
This implies that \(\theta = 90^\circ\). Therefore, the line is perpendicular to the normal of the plane, which means it is parallel to the plane.

### Step 3: Find a point on the line.
We can find a point on the line by substituting \(t = 0\):
- \(x = 2\)
- \(y = 1\)
- \(z = 1\)

Thus, the point \(P(2, 1, 1)\) lies on the line.

### Step 4: Find the image of the point in the plane.
To find the image of the point \(P(2, 1, 1)\) in the plane \(x + y + z = 7\), we can use the formula for the reflection of a point across a plane.

Let the image point be \(Q(\alpha, \beta, \gamma)\). The point \(Q\) must satisfy the equation of the plane:
\[ \alpha + \beta + \gamma = 7 \]
The line connecting \(P\) and \(Q\) must be perpendicular to the plane, which means it must be in the direction of the normal vector \((1, 1, 1)\).

Using the formula for the reflection:
\[ Q = P + 2 \cdot \text{distance from } P \text{ to the plane} \cdot \text{unit normal vector} \]
The distance from point \(P(2, 1, 1)\) to the plane can be calculated as:
\[ \text{Distance} = \frac{|2 + 1 + 1 - 7|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|4 - 7|}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]
Thus, the image point \(Q\) can be calculated as:
\[ Q = (2, 1, 1) + 2\sqrt{3} \cdot \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = (2 + 2, 1 + 2, 1 + 2) = (4, 3, 3) \]

### Step 5: Write the equation of the image line.
The image line will have the same direction ratios as the original line, which are \(3, 1, -4\), and it passes through the point \(Q(4, 3, 3)\). Therefore, the equation of the image line is:
\[ \frac{x - 4}{3} = \frac{y - 3}{1} = \frac{z - 3}{-4} \]

### Final Answer:
The image of the line in the plane \(x + y + z = 7\) is given by:
\[ \frac{x - 4}{3} = \frac{y - 3}{1} = \frac{z - 3}{-4} \]