The acute angle between the two lines with direction ratios `(1,-2,-2) and (2,-2,1)` is

Options

A. `cos^(-1)(2/3)`
B. `cos^(-1)(4/9)`
C. `cos^(-1)(2/9)`
D. `cos^(-1)(5/9)`

Explanation

To find the acute angle between the two lines with direction ratios (1, -2, -2) and (2, -2, 1), we can follow these steps:

### Step 1: Define the direction vectors
Let the direction ratios of the first line be represented as vector **b1** and the second line as vector **b2**.

\[ \mathbf{b_1} = (1, -2, -2) \quad \text{and} \quad \mathbf{b_2} = (2, -2, 1) \]

### Step 2: Calculate the dot product of the two vectors
The dot product of two vectors **a** and **b** is given by:

\[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]

For our vectors **b1** and **b2**:

\[ \mathbf{b_1} \cdot \mathbf{b_2} = (1)(2) + (-2)(-2) + (-2)(1) \]

Calculating this gives:

\[ = 2 + 4 - 2 = 4 \]

### Step 3: Calculate the magnitudes of the vectors
The magnitude of a vector **a** is given by:

\[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

Calculating the magnitude of **b1**:

\[ |\mathbf{b_1}| = \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Calculating the magnitude of **b2**:

\[ |\mathbf{b_2}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

### Step 4: Use the dot product to find the cosine of the angle
The cosine of the angle θ between two vectors is given by:

\[ \cos \theta = \frac{\mathbf{b_1} \cdot \mathbf{b_2}}{|\mathbf{b_1}| |\mathbf{b_2}|} \]

Substituting the values we calculated:

\[ \cos \theta = \frac{4}{3 \cdot 3} = \frac{4}{9} \]

### Step 5: Calculate the angle θ
To find the angle θ, we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{4}{9}\right) \]

This gives us the acute angle between the two lines.

### Final Result
Thus, the acute angle between the two lines is:

\[ \theta = \cos^{-1}\left(\frac{4}{9}\right) \]

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