To solve the problem of finding the projection of the line segment AB on a line that makes specific angles with the coordinate axes, we will follow these steps:
### Step 1: Find the Direction Ratios of Line AB
Given points A(-1, 2, 1) and B(4, 3, 5), we first determine the direction ratios of the line segment AB.
\[
\text{Direction Ratios of } AB = (4 - (-1), 3 - 2, 5 - 1) = (5, 1, 4)
\]
### Step 2: Determine the Direction Cosines of the Given Line
The angles made by the line with the Y and Z axes are given as 120° and 135°, respectively. We need to find the direction cosines corresponding to these angles.
1. **For Y-axis (120°)**:
\[
\cos(120°) = -\cos(60°) = -\frac{1}{2}
\]
2. **For Z-axis (135°)**:
\[
\cos(135°) = -\cos(45°) = -\frac{1}{\sqrt{2}}
\]
Let \(\gamma\) be the angle with the X-axis. The relationship between the direction cosines is given by:
\[
\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1
\]
Substituting the known values:
\[
\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \gamma = 1
\]
\[
\frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1
\]
\[
\frac{1}{4} + \frac{2}{4} + \cos^2 \gamma = 1
\]
\[
\frac{3}{4} + \cos^2 \gamma = 1
\]
\[
\cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}
\]
Thus,
\[
\cos \gamma = \frac{1}{2} \quad (\text{since } \gamma \text{ is acute})
\]
### Step 3: Find the Direction Cosines of the Line
Now we have the direction cosines:
- \(\cos \alpha = -\frac{1}{2}\)
- \(\cos \beta = -\frac{1}{\sqrt{2}}\)
- \(\cos \gamma = \frac{1}{2}\)
### Step 4: Form the Direction Vector of the Line
The direction vector of the line can be expressed as:
\[
\mathbf{c} = k \left(-\frac{1}{2}, -\frac{1}{\sqrt{2}}, \frac{1}{2}\right)
\]
where \(k\) is a positive scalar.
### Step 5: Calculate the Projection of AB on the Line
The projection of vector \(\mathbf{AB}\) onto vector \(\mathbf{c}\) is given by:
\[
\text{Projection} = \frac{\mathbf{AB} \cdot \mathbf{c}}{|\mathbf{c}|}
\]
First, we find the magnitude of \(\mathbf{c}\):
\[
|\mathbf{c}| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{2} + \frac{1}{4}} = \sqrt{1} = 1
\]
Next, we compute the dot product \(\mathbf{AB} \cdot \mathbf{c}\):
\[
\mathbf{AB} = (5, 1, 4)
\]
\[
\mathbf{c} = \left(-\frac{1}{2}, -\frac{1}{\sqrt{2}}, \frac{1}{2}\right)
\]
\[
\mathbf{AB} \cdot \mathbf{c} = 5 \left(-\frac{1}{2}\right) + 1 \left(-\frac{1}{\sqrt{2}}\right) + 4 \left(\frac{1}{2}\right)
\]
\[
= -\frac{5}{2} - \frac{1}{\sqrt{2}} + 2 = -\frac{5}{2} + 2 - \frac{1}{\sqrt{2}} = -\frac{1}{2} - \frac{1}{\sqrt{2}}
\]
### Step 6: Final Projection Calculation
Thus, the projection of AB on the line is:
\[
\text{Projection} = -\frac{1}{2} - \frac{1}{\sqrt{2}} \text{ units}
\]
