Let `A(-1, 2, 1) and B(4, 3, 5)` be two given points. Find the projection of AB on a line which makes angle `120^(@) and 135^(@)` with Yand Z-axes respectively, and an acute angle with X-axis.

Answer

To find the projection of the line segment \( AB \) on a line that makes angles of \( 120^\circ \) with the Y-axis and \( 135^\circ \) with the Z-axis, we can follow these steps:

### Step 1: Determine the coordinates of points A and B
Given points are:
- \( A(-1, 2, 1) \)
- \( B(4, 3, 5) \)

### Step 2: Calculate the direction vector \( \overrightarrow{AB} \)
The direction vector \( \overrightarrow{AB} \) is calculated as:
\[ \overrightarrow{AB} = B - A = (4 - (-1), 3 - 2, 5 - 1) = (5, 1, 4) \]

### Step 3: Find the direction cosines for the line
The angles given are:
- \( \beta = 120^\circ \) (with Y-axis)
- \( \gamma = 135^\circ \) (with Z-axis)

Using the cosine values:
\[ \cos \beta = \cos(120^\circ) = -\frac{1}{2} \]
\[ \cos \gamma = \cos(135^\circ) = -\frac{1}{\sqrt{2}} \]

### Step 4: Use the relation for direction cosines
The relationship for direction cosines is:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]
Substituting the values we have:
\[ \cos^2 \alpha + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 = 1 \]
\[ \cos^2 \alpha + \frac{1}{4} + \frac{1}{2} = 1 \]
\[ \cos^2 \alpha + \frac{3}{4} = 1 \]
\[ \cos^2 \alpha = 1 - \frac{3}{4} = \frac{1}{4} \]
Thus,
\[ \cos \alpha = \pm \frac{1}{2} \]
Since the angle with the X-axis is acute, we take the positive value:
\[ \cos \alpha = \frac{1}{2} \]

### Step 5: Write the direction vector of the line
The direction cosines give us the direction vector:
\[ \mathbf{v} = (\cos \alpha, \cos \beta, \cos \gamma) = \left(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}}\right) \]

### Step 6: Calculate the projection of \( \overrightarrow{AB} \) on \( \mathbf{v} \)
The projection of vector \( \overrightarrow{AB} \) on vector \( \mathbf{v} \) is given by:
\[ \text{Projection} = \frac{\overrightarrow{AB} \cdot \mathbf{v}}{|\mathbf{v}|} \]
First, calculate the dot product \( \overrightarrow{AB} \cdot \mathbf{v} \):
\[ \overrightarrow{AB} \cdot \mathbf{v} = (5, 1, 4) \cdot \left(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}}\right) \]
\[ = 5 \cdot \frac{1}{2} + 1 \cdot \left(-\frac{1}{2}\right) + 4 \cdot \left(-\frac{1}{\sqrt{2}}\right) \]
\[ = \frac{5}{2} - \frac{1}{2} - \frac{4}{\sqrt{2}} \]
\[ = 2 - \frac{4\sqrt{2}}{2} = 2 - 2\sqrt{2} \]

### Step 7: Calculate the magnitude of \( \mathbf{v} \)
The magnitude of \( \mathbf{v} \) is:
\[ |\mathbf{v}| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} \]
\[ = \sqrt{\frac{1}{4} + \frac{1}{4} + \frac{1}{2}} = \sqrt{1} = 1 \]

### Step 8: Final projection calculation
Thus, the projection of \( \overrightarrow{AB} \) on the line is:
\[ \text{Projection} = 2 - 2\sqrt{2} \]

### Final Answer
The projection of line segment \( AB \) on the specified line is \( 2 - 2\sqrt{2} \).