The acute angle between the lines whose direction ratios are 1, 1, 2 and `sqrt(3)-1, -sqrt(3)-1,4` is

Options

A. `45^(@)`
B. `30^(@)`
C. `90^(@)`
D. `60^(@)`

Answer

To find the acute angle between the two lines given by their direction ratios, we can follow these steps:

### Step 1: Identify the Direction Ratios
The direction ratios of the two lines are given as:
- Line 1: \( (1, 1, 2) \)
- Line 2: \( (\sqrt{3} - 1, -\sqrt{3} - 1, 4) \)

### Step 2: Represent the Direction Ratios as Vectors
We can represent these direction ratios as vectors:
- Let \( \mathbf{a} = \langle 1, 1, 2 \rangle \)
- Let \( \mathbf{b} = \langle \sqrt{3} - 1, -\sqrt{3} - 1, 4 \rangle \)

### Step 3: Calculate the Dot Product
The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = (1)(\sqrt{3} - 1) + (1)(-\sqrt{3} - 1) + (2)(4) \]
Calculating each term:
- First term: \( 1 \cdot (\sqrt{3} - 1) = \sqrt{3} - 1 \)
- Second term: \( 1 \cdot (-\sqrt{3} - 1) = -\sqrt{3} - 1 \)
- Third term: \( 2 \cdot 4 = 8 \)

Now, summing these:
\[ \mathbf{a} \cdot \mathbf{b} = (\sqrt{3} - 1) + (-\sqrt{3} - 1) + 8 = 6 \]

### Step 4: Calculate the Magnitudes of the Vectors
Now we need to calculate the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \).

For \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

For \( \mathbf{b} \):
\[ |\mathbf{b}| = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2} \]
Calculating \( (\sqrt{3} - 1)^2 \):
\[ (\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \]
Calculating \( (-\sqrt{3} - 1)^2 \):
\[ (-\sqrt{3} - 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \]
Now summing these:
\[ |\mathbf{b}|^2 = (4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) + 16 = 24 \]
Thus,
\[ |\mathbf{b}| = \sqrt{24} = 2\sqrt{6} \]

### Step 5: Use the Dot Product to Find Cosine of the Angle
Using the formula for the cosine of the angle \( \theta \) between two vectors:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]
Substituting the values we found:
\[ \cos \theta = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{12} = \frac{1}{2} \]

### Step 6: Find the Angle
Now, we find the angle \( \theta \):
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Final Answer
The acute angle between the lines is \( 60^\circ \).

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