The length of projection of the line segment joining the points `(1,0,-1) and (-1,2,2)` on the plane `x+3y-5z=6` is equal to

Options

A. 2
B. `sqrt((271)/(53))`
C. `sqrt((472)/(31))`
D. `sqrt((474)/(35))`

Explanation

To find the length of the projection of the line segment joining the points \( A(1, 0, -1) \) and \( B(-1, 2, 2) \) on the plane given by the equation \( x + 3y - 5z = 6 \), we can follow these steps:

### Step 1: Find the coordinates of points A and B
Let:
- \( A = (1, 0, -1) \)
- \( B = (-1, 2, 2) \)

### Step 2: Calculate the distance between points A and B
Using the distance formula for points in 3D:
\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates of points A and B:
\[ D = \sqrt{((-1) - 1)^2 + (2 - 0)^2 + (2 - (-1))^2} \]
\[ D = \sqrt{(-2)^2 + (2)^2 + (3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \]

### Step 3: Find the direction vector \( \overrightarrow{AB} \)
The direction vector \( \overrightarrow{AB} \) is given by:
\[ \overrightarrow{AB} = B - A = (-1 - 1, 2 - 0, 2 - (-1)) = (-2, 2, 3) \]

### Step 4: Identify the normal vector of the plane
The equation of the plane is \( x + 3y - 5z = 6 \). The normal vector \( \mathbf{n} \) to the plane can be extracted from the coefficients of \( x, y, z \):
\[ \mathbf{n} = (1, 3, -5) \]

### Step 5: Calculate the dot product \( \mathbf{n} \cdot \overrightarrow{AB} \)
\[ \mathbf{n} \cdot \overrightarrow{AB} = (1)(-2) + (3)(2) + (-5)(3) = -2 + 6 - 15 = -11 \]

### Step 6: Calculate the magnitudes of \( \mathbf{n} \) and \( \overrightarrow{AB} \)
Magnitude of \( \mathbf{n} \):
\[ |\mathbf{n}| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \]

Magnitude of \( \overrightarrow{AB} \):
\[ |\overrightarrow{AB}| = \sqrt{(-2)^2 + (2)^2 + (3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17} \]

### Step 7: Calculate \( \cos \theta \)
Using the formula for the cosine of the angle between two vectors:
\[ \cos \theta = \frac{\mathbf{n} \cdot \overrightarrow{AB}}{|\mathbf{n}| |\overrightarrow{AB}|} \]
Substituting the values:
\[ \cos \theta = \frac{-11}{\sqrt{35} \cdot \sqrt{17}} = \frac{-11}{\sqrt{595}} \]

### Step 8: Calculate \( \sin \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^2 \theta = 1 - \left(\frac{-11}{\sqrt{595}}\right)^2 = 1 - \frac{121}{595} = \frac{474}{595} \]
Thus,
\[ \sin \theta = \sqrt{\frac{474}{595}} = \frac{\sqrt{474}}{\sqrt{595}} \]

### Step 9: Calculate the length of the projection
The length of the projection of \( \overrightarrow{AB} \) on the plane is given by:
\[ \text{Length of projection} = |\overrightarrow{AB}| \cdot \sin \theta \]
Substituting the values:
\[ \text{Length of projection} = \sqrt{17} \cdot \frac{\sqrt{474}}{\sqrt{595}} = \frac{\sqrt{17 \cdot 474}}{\sqrt{595}} = \frac{\sqrt{8058}}{\sqrt{595}} \]

### Final Result
The length of the projection of the line segment joining the points \( (1,0,-1) \) and \( (-1,2,2) \) on the plane \( x + 3y - 5z = 6 \) is:
\[ \frac{\sqrt{8058}}{\sqrt{595}} \]