Three lines with direction cosines `(1,1,2),(sqrt(3)-1,-sqrt(3)-1,4), (sqrt(3)-1,sqrt(3)-1,4)` enclose

Options

A. an isosceles triangle
B. a right angled tringle
C. an equilateral tringle
D. a right angled isosceles triangle

Answer

To solve the problem of determining the type of triangle formed by the three given lines with direction cosines, we will follow these steps:

### Step 1: Identify the Direction Cosines
The direction cosines of the three lines are given as:
- Line 1 (L1): (1, 1, 2)
- Line 2 (L2): (√3 - 1, -√3 - 1, 4)
- Line 3 (L3): (√3 - 1, √3 - 1, 4)

### Step 2: Calculate the Lengths of the Line Segments
To determine the type of triangle formed by these lines, we need to calculate the distances between each pair of lines using the distance formula in three-dimensional space.

The distance formula between two points \( P_1(x_1, y_1, z_1) \) and \( P_2(x_2, y_2, z_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

### Step 3: Calculate the Distance L1 to L2
Using the coordinates of L1 and L2:
- L1: (1, 1, 2)
- L2: (√3 - 1, -√3 - 1, 4)

Calculating the distance \( L1 \) to \( L2 \):
\[ d_{L1L2} = \sqrt{((\sqrt{3} - 1) - 1)^2 + ((-\sqrt{3} - 1) - 1)^2 + (4 - 2)^2} \]
\[ = \sqrt{(\sqrt{3} - 2)^2 + (-\sqrt{3} - 2)^2 + 2^2} \]
Calculating each term:
- \( (\sqrt{3} - 2)^2 = 3 - 4\sqrt{3} + 4 = 7 - 4\sqrt{3} \)
- \( (-\sqrt{3} - 2)^2 = 3 + 4\sqrt{3} + 4 = 7 + 4\sqrt{3} \)
- \( 2^2 = 4 \)

Putting it all together:
\[ d_{L1L2} = \sqrt{(7 - 4\sqrt{3}) + (7 + 4\sqrt{3}) + 4} = \sqrt{18} = 3\sqrt{2} \]

### Step 4: Calculate the Distance L2 to L3
Using the coordinates of L2 and L3:
- L2: (√3 - 1, -√3 - 1, 4)
- L3: (√3 - 1, √3 - 1, 4)

Calculating the distance \( L2 \) to \( L3 \):
\[ d_{L2L3} = \sqrt{((\sqrt{3} - 1) - (\sqrt{3} - 1))^2 + ((-\sqrt{3} - 1) - (\sqrt{3} - 1))^2 + (4 - 4)^2} \]
\[ = \sqrt{0 + (-2\sqrt{3})^2 + 0} = \sqrt{12} = 2\sqrt{3} \]

### Step 5: Calculate the Distance L1 to L3
Using the coordinates of L1 and L3:
- L1: (1, 1, 2)
- L3: (√3 - 1, √3 - 1, 4)

Calculating the distance \( L1 \) to \( L3 \):
\[ d_{L1L3} = \sqrt{((\sqrt{3} - 1) - 1)^2 + ((\sqrt{3} - 1) - 1)^2 + (4 - 2)^2} \]
This is similar to the calculation for L1 to L2:
\[ = \sqrt{(\sqrt{3} - 2)^2 + (\sqrt{3} - 2)^2 + 2^2} \]
\[ = \sqrt{2(7 - 4\sqrt{3}) + 4} = \sqrt{18} = 3\sqrt{2} \]

### Step 6: Compare the Distances
Now we have the distances:
- \( d_{L1L2} = 3\sqrt{2} \)
- \( d_{L2L3} = 2\sqrt{3} \)
- \( d_{L1L3} = 3\sqrt{2} \)

### Step 7: Determine the Type of Triangle
Since \( d_{L1L2} = d_{L1L3} \), the triangle formed by these lines is an isosceles triangle.

### Conclusion
The three lines with the given direction cosines enclose an isosceles triangle.

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