The equation of the projection line of the line `(x+1)/(2)=(y+1)/(-1)=(z+3)/(4)` on the plane `x+2y+z=6` is

Options

A. `(x-1)/(4)=(y-3)/(7)=(z-1)/(10)`
B. `(x-1)/(-4)=(y+3)/(7)=(z-1)/(10)`
C. `(x-1)/(4)=(y-3)/(-7)=(z+1)/(10)`
D. `(x+3)/(4)=(y-2)/(7)=(z-7)/(-10)`

Answer

To find the equation of the projection line of the given line on the specified plane, we can follow these steps:

### Step 1: Identify the given line and plane
The line is given in symmetric form:
\[ \frac{x+1}{2} = \frac{y+1}{-1} = \frac{z+3}{4} \]
The plane is given by the equation:
\[ x + 2y + z = 6 \]

### Step 2: Parametrize the line
Let \( \lambda \) be the parameter. We can express the coordinates \( x, y, z \) in terms of \( \lambda \):
\[ x = 2\lambda - 1, \quad y = -\lambda - 1, \quad z = 4\lambda - 3 \]

### Step 3: Substitute into the plane equation
We need to find the value of \( \lambda \) for which the point on the line lies on the plane. Substitute \( x, y, z \) into the plane equation:
\[ (2\lambda - 1) + 2(-\lambda - 1) + (4\lambda - 3) = 6 \]

### Step 4: Simplify the equation
Now simplify the equation:
\[ 2\lambda - 1 - 2\lambda - 2 + 4\lambda - 3 = 6 \]
Combine like terms:
\[ 4\lambda - 6 = 6 \]
\[ 4\lambda = 12 \]
\[ \lambda = 3 \]

### Step 5: Find the coordinates of point A
Now substitute \( \lambda = 3 \) back into the parametric equations to find the coordinates of point A:
\[ x = 2(3) - 1 = 5, \quad y = -3 - 1 = -4, \quad z = 4(3) - 3 = 9 \]
Thus, point A is \( (5, -4, 9) \).

### Step 6: Find point B (the foot of the perpendicular)
The normal vector to the plane \( x + 2y + z = 6 \) is given by the coefficients of \( x, y, z \), which are \( (1, 2, 1) \). The direction ratios of the line perpendicular to the plane are \( (1, 2, 1) \).

Using point A and the direction ratios, we can find point B by moving along the normal vector from point A:
Let point B be \( (x_B, y_B, z_B) \):
\[ x_B = 5 + t \cdot 1, \quad y_B = -4 + t \cdot 2, \quad z_B = 9 + t \cdot 1 \]
Substituting into the plane equation:
\[ (5 + t) + 2(-4 + 2t) + (9 + t) = 6 \]
Simplifying:
\[ 5 + t - 8 + 4t + 9 + t = 6 \]
\[ 6t + 6 = 6 \]
\[ 6t = 0 \implies t = 0 \]
Thus, point B is the same as point A, \( (5, -4, 9) \).

### Step 7: Find the direction ratios of the projection line
The projection line will be in the direction of the normal vector of the plane, which is \( (1, 2, 1) \).

### Step 8: Write the equation of the projection line
Using point A and the direction ratios, the equation of the projection line can be expressed as:
\[ \frac{x - 5}{1} = \frac{y + 4}{2} = \frac{z - 9}{1} \]

### Final Answer
The equation of the projection line is:
\[ \frac{x - 5}{1} = \frac{y + 4}{2} = \frac{z - 9}{1} \]

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