To find the equations of the projection of the line given by \((x+1)/-2 = (y-1)/3 = (z+2)/4\) onto the plane \(2x + y + 4z = 1\), we will follow these steps:
### Step 1: Identify the line's parametric equations
The line can be expressed in parametric form. Let \(t\) be the parameter. We can rewrite the line as:
\[
\frac{x + 1}{-2} = \frac{y - 1}{3} = \frac{z + 2}{4} = t
\]
From this, we can derive the parametric equations:
\[
x = -2t - 1, \quad y = 3t + 1, \quad z = 4t - 2
\]
### Step 2: Substitute the parametric equations into the plane equation
We substitute \(x\), \(y\), and \(z\) into the plane equation \(2x + y + 4z = 1\):
\[
2(-2t - 1) + (3t + 1) + 4(4t - 2) = 1
\]
Expanding this gives:
\[
-4t - 2 + 3t + 1 + 16t - 8 = 1
\]
Combining like terms:
\[
(15t - 9) = 1
\]
Solving for \(t\):
\[
15t = 10 \implies t = \frac{2}{3}
\]
### Step 3: Find the coordinates of point B on the line
Now we substitute \(t = \frac{2}{3}\) back into the parametric equations to find the coordinates of point \(B\):
\[
x_B = -2\left(\frac{2}{3}\right) - 1 = -\frac{4}{3} - 1 = -\frac{7}{3}
\]
\[
y_B = 3\left(\frac{2}{3}\right) + 1 = 2 + 1 = 3
\]
\[
z_B = 4\left(\frac{2}{3}\right) - 2 = \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}
\]
Thus, the coordinates of point \(B\) are:
\[
B\left(-\frac{7}{3}, 3, \frac{2}{3}\right)
\]
### Step 4: Find the normal vector to the plane
The normal vector \(\mathbf{n}\) to the plane \(2x + y + 4z = 1\) is given by the coefficients of \(x\), \(y\), and \(z\):
\[
\mathbf{n} = (2, 1, 4)
\]
### Step 5: Write the equation of the normal line
The normal line passes through point \(B\) and is directed along the normal vector \(\mathbf{n}\). The parametric equations of the normal line can be written as:
\[
x = -\frac{7}{3} + 2s, \quad y = 3 + s, \quad z = \frac{2}{3} + 4s
\]
### Step 6: Find the intersection of the normal line with the plane
To find the projection point \(A'\) on the plane, we substitute the parametric equations of the normal line into the plane equation:
\[
2\left(-\frac{7}{3} + 2s\right) + (3 + s) + 4\left(\frac{2}{3} + 4s\right) = 1
\]
Expanding this gives:
\[
-\frac{14}{3} + 4s + 3 + s + \frac{8}{3} + 16s = 1
\]
Combining like terms:
\[
(21s) + \left(-\frac{14}{3} + 3 + \frac{8}{3}\right) = 1
\]
\[
21s - \frac{6}{3} = 1 \implies 21s - 2 = 1 \implies 21s = 3 \implies s = \frac{1}{7}
\]
### Step 7: Find the coordinates of point \(A'\)
Substituting \(s = \frac{1}{7}\) back into the equations of the normal line:
\[
x_{A'} = -\frac{7}{3} + 2\left(\frac{1}{7}\right) = -\frac{7}{3} + \frac{2}{7}
\]
Finding a common denominator (21):
\[
x_{A'} = -\frac{49}{21} + \frac{6}{21} = -\frac{43}{21}
\]
\[
y_{A'} = 3 + \frac{1}{7} = \frac{21}{7} + \frac{1}{7} = \frac{22}{7}
\]
\[
z_{A'} = \frac{2}{3} + 4\left(\frac{1}{7}\right) = \frac{2}{3} + \frac{4}{7}
\]
Finding a common denominator (21):
\[
z_{A'} = \frac{14}{21} + \frac{12}{21} = \frac{26}{21}
\]
Thus, the coordinates of point \(A'\) are:
\[
A'\left(-\frac{43}{21}, \frac{22}{7}, \frac{26}{21}\right)
\]
### Step 8: Write the equation of the projection line
Now, we can find the equation of the line passing through points \(A'\) and \(B\):
Using the two points, we can write the equation:
\[
\frac{x + \frac{7}{3}}{-\frac{43}{21} + \frac{7}{3}} = \frac{y - 3}{\frac{22}{7} - 3} = \frac{z - \frac{2}{3}}{\frac{26}{21} - \frac{2}{3}}
\]
### Final Result
The projection of the line onto the plane is given by the above equation.
