The angle between the lines
`(x-1)/(2)=(y-2)/(1)=(z+3)/(2)` and `(x)/(1)=(y)/(1)=(z)/(0)` is

Options

A. `0^(@)`
B. `30^(@)`
C. `45^(@)`
D. `90^(@)`

Answer

To find the angle between the given lines, we will follow these steps:

### Step 1: Identify the direction vectors of the lines.
The first line is given by:
\[ \frac{x-1}{2} = \frac{y-2}{1} = \frac{z+3}{2} \]
From this, we can extract the direction vector \( \mathbf{D_1} \):
\[ \mathbf{D_1} = \langle 2, 1, 2 \rangle \]

The second line is given by:
\[ \frac{x}{1} = \frac{y}{1} = \frac{z}{0} \]
From this, we can extract the direction vector \( \mathbf{D_2} \):
\[ \mathbf{D_2} = \langle 1, 1, 0 \rangle \]

### Step 2: Use the dot product formula to find the angle.
The formula for the angle \( \theta \) between two vectors \( \mathbf{D_1} \) and \( \mathbf{D_2} \) is given by:
\[ \mathbf{D_1} \cdot \mathbf{D_2} = |\mathbf{D_1}| |\mathbf{D_2}| \cos \theta \]

### Step 3: Calculate the dot product \( \mathbf{D_1} \cdot \mathbf{D_2} \).
\[ \mathbf{D_1} \cdot \mathbf{D_2} = (2)(1) + (1)(1) + (2)(0) = 2 + 1 + 0 = 3 \]

### Step 4: Calculate the magnitudes of \( \mathbf{D_1} \) and \( \mathbf{D_2} \).
\[ |\mathbf{D_1}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]
\[ |\mathbf{D_2}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 5: Substitute into the formula to find \( \cos \theta \).
\[ \cos \theta = \frac{\mathbf{D_1} \cdot \mathbf{D_2}}{|\mathbf{D_1}| |\mathbf{D_2}|} = \frac{3}{3 \cdot \sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 6: Find the angle \( \theta \).
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \]

### Conclusion
The angle between the two lines is \( 45^\circ \).

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