To find the image of the line given by the equations \((x-1)/2 = (y+1)/(-1) = (z-3)/4\) in the plane defined by the equation \(3x - 3y + 10z - 26 = 0\), we will follow these steps:
### Step 1: Identify the line's point and direction ratios
The line can be expressed in parametric form. We can set:
\[
\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4} = t
\]
From this, we can derive the parametric equations:
- \(x = 2t + 1\)
- \(y = -t - 1\)
- \(z = 4t + 3\)
The point on the line when \(t = 0\) is \(P(1, -1, 3)\) and the direction ratios of the line are \( (2, -1, 4) \).
### Step 2: Write the equation of the plane
The equation of the plane is given as:
\[
3x - 3y + 10z - 26 = 0
\]
We can rewrite this in the standard form:
\[
3x - 3y + 10z = 26
\]
### Step 3: Find the normal vector of the plane
The normal vector \( \mathbf{n} \) of the plane can be derived from the coefficients of \(x\), \(y\), and \(z\):
\[
\mathbf{n} = (3, -3, 10)
\]
### Step 4: Find the foot of the perpendicular from point P to the plane
To find the foot of the perpendicular from point \(P(1, -1, 3)\) to the plane, we can use the formula for the foot of the perpendicular:
\[
\text{Foot} = P + \lambda \mathbf{n}
\]
where \(\lambda\) is a scalar that we need to determine.
### Step 5: Substitute the foot into the plane equation
Let the foot of the perpendicular be \(F(x, y, z)\):
\[
F(1 + 3\lambda, -1 - 3\lambda, 3 + 10\lambda)
\]
Substituting these coordinates into the plane equation:
\[
3(1 + 3\lambda) - 3(-1 - 3\lambda) + 10(3 + 10\lambda) = 26
\]
Expanding this gives:
\[
3 + 9\lambda + 3 + 9\lambda + 30 + 100\lambda = 26
\]
Combining like terms:
\[
12 + 118\lambda = 26
\]
Solving for \(\lambda\):
\[
118\lambda = 26 - 12 = 14 \implies \lambda = \frac{14}{118} = \frac{7}{59}
\]
### Step 6: Find the coordinates of the foot of the perpendicular
Now substituting \(\lambda\) back into \(F\):
\[
F\left(1 + 3\left(\frac{7}{59}\right), -1 - 3\left(\frac{7}{59}\right), 3 + 10\left(\frac{7}{59}\right)\right)
\]
Calculating each coordinate:
- \(x = 1 + \frac{21}{59} = \frac{59 + 21}{59} = \frac{80}{59}\)
- \(y = -1 - \frac{21}{59} = -\frac{59 + 21}{59} = -\frac{80}{59}\)
- \(z = 3 + \frac{70}{59} = \frac{177 + 70}{59} = \frac{247}{59}\)
Thus, the foot of the perpendicular \(F\) is:
\[
F\left(\frac{80}{59}, -\frac{80}{59}, \frac{247}{59}\right)
\]
### Step 7: Find the image of the point
The image \(I\) of point \(P\) with respect to the plane can be found using:
\[
I = P + 2(\text{Foot} - P)
\]
Calculating this gives:
\[
I = P + 2\left(F - P\right)
\]
Substituting \(P(1, -1, 3)\) and \(F\):
\[
I = \left(1, -1, 3\right) + 2\left(\left(\frac{80}{59}, -\frac{80}{59}, \frac{247}{59}\right) - \left(1, -1, 3\right)\right)
\]
Calculating \(F - P\):
\[
F - P = \left(\frac{80}{59} - 1, -\frac{80}{59} + 1, \frac{247}{59} - 3\right) = \left(\frac{80 - 59}{59}, \frac{-80 + 59}{59}, \frac{247 - 177}{59}\right) = \left(\frac{21}{59}, -\frac{21}{59}, \frac{70}{59}\right)
\]
Now substituting back:
\[
I = \left(1, -1, 3\right) + 2\left(\frac{21}{59}, -\frac{21}{59}, \frac{70}{59}\right) = \left(1 + \frac{42}{59}, -1 - \frac{42}{59}, 3 + \frac{140}{59}\right)
\]
Calculating the final coordinates:
- \(x = \frac{59 + 42}{59} = \frac{101}{59}\)
- \(y = -\frac{59 + 42}{59} = -\frac{101}{59}\)
- \(z = \frac{177 + 140}{59} = \frac{317}{59}\)
Thus, the image of the line in the plane is given by the point:
\[
I\left(\frac{101}{59}, -\frac{101}{59}, \frac{317}{59}\right)
\]
