If direction-cosines of two lines are proportional to 4,3,2 and 1, -2, 1, then the angle between the lines is :

Options

A. `90^(@)`
B. `60^(@)`
C. `45^(@)`
D. None of these.

Answer

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

### Step 1: Identify the direction ratios
The direction cosines of the first line (L1) are proportional to \(4, 3, 2\) and for the second line (L2) are proportional to \(1, -2, 1\). We can take the direction ratios as:
- For L1: \( (4, 3, 2) \)
- For L2: \( (1, -2, 1) \)

### Step 2: Use the formula for the cosine of the angle between two lines
The cosine of the angle \( \theta \) between two lines can be calculated using the formula:
\[ \cos \theta = \frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}} \]
where \( (A_1, B_1, C_1) \) and \( (A_2, B_2, C_2) \) are the direction ratios of the two lines.

### Step 3: Substitute the values into the formula
Substituting the values from our direction ratios:
- \( A_1 = 4, B_1 = 3, C_1 = 2 \)
- \( A_2 = 1, B_2 = -2, C_2 = 1 \)

Calculating the numerator:
\[ A_1 A_2 + B_1 B_2 + C_1 C_2 = 4 \cdot 1 + 3 \cdot (-2) + 2 \cdot 1 = 4 - 6 + 2 = 0 \]

Calculating the denominator:
\[ \sqrt{A_1^2 + B_1^2 + C_1^2} = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29} \]
\[ \sqrt{A_2^2 + B_2^2 + C_2^2} = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]

Thus, the denominator becomes:
\[ \sqrt{29} \cdot \sqrt{6} = \sqrt{174} \]

### Step 4: Calculate \( \cos \theta \)
Now substituting back into the formula:
\[ \cos \theta = \frac{0}{\sqrt{174}} = 0 \]

### Step 5: Determine the angle \( \theta \)
Since \( \cos \theta = 0 \), we have:
\[ \theta = 90^\circ \text{ or } \frac{\pi}{2} \text{ radians} \]

### Final Answer
The angle between the two lines is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.

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