What is the acute angle between the pair of straight line `sqrt2x + sqrt3y = 1` and `sqrt3x + sqrt2y = 2` ?

Options

A. `tan^(-1) ((1)/(2sqrt6))`
B. `tan^(-1) ((1)/(sqrt2))`
C. `tan^(-1) (3)`
D. `tan^(-1) ((1)/(sqrt3))`

Answer

To find the acute angle between the two straight lines given by the equations \( \sqrt{2}x + \sqrt{3}y = 1 \) and \( \sqrt{3}x + \sqrt{2}y = 2 \), we can follow these steps:

### Step 1: Write the equations in slope-intercept form
We need to convert both equations into the form \( y = mx + c \), where \( m \) is the slope.

1. **For the first equation**:
\[ \sqrt{2}x + \sqrt{3}y = 1 \]
Rearranging gives:
\[ \sqrt{3}y = 1 - \sqrt{2}x \]
Dividing by \( \sqrt{3} \):
\[ y = -\frac{\sqrt{2}}{\sqrt{3}}x + \frac{1}{\sqrt{3}} \]
Thus, the slope \( m_1 = -\frac{\sqrt{2}}{\sqrt{3}} \).

2. **For the second equation**:
\[ \sqrt{3}x + \sqrt{2}y = 2 \]
Rearranging gives:
\[ \sqrt{2}y = 2 - \sqrt{3}x \]
Dividing by \( \sqrt{2} \):
\[ y = -\frac{\sqrt{3}}{\sqrt{2}}x + \frac{2}{\sqrt{2}} \]
Thus, the slope \( m_2 = -\frac{\sqrt{3}}{\sqrt{2}} \).

### Step 2: Use the formula for the angle between two lines
The formula for the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

### Step 3: Substitute the slopes into the formula
Substituting \( m_1 \) and \( m_2 \):
\[ \tan(\theta) = \left| \frac{-\frac{\sqrt{2}}{\sqrt{3}} + \frac{\sqrt{3}}{\sqrt{2}}}{1 + \left(-\frac{\sqrt{2}}{\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{\sqrt{2}}\right)} \right| \]

Calculating the numerator:
\[ -\frac{\sqrt{2}}{\sqrt{3}} + \frac{\sqrt{3}}{\sqrt{2}} = \frac{-\sqrt{2}\sqrt{2} + \sqrt{3}\sqrt{3}}{\sqrt{6}} = \frac{-2 + 3}{\sqrt{6}} = \frac{1}{\sqrt{6}} \]

Calculating the denominator:
\[ 1 + \left(-\frac{\sqrt{2}}{\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{\sqrt{2}}\right) = 1 + 1 = 2 \]

Thus:
\[ \tan(\theta) = \left| \frac{\frac{1}{\sqrt{6}}}{2} \right| = \frac{1}{2\sqrt{6}} \]

### Step 4: Find the angle \( \theta \)
To find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{1}{2\sqrt{6}}\right) \]

### Final Answer
The acute angle between the two lines is:
\[ \theta = \tan^{-1}\left(\frac{1}{2\sqrt{6}}\right) \]