To find the value of [tex]\( k \)[/tex] for which the acute angle between the lines [tex]\( 2x + y = 5 \)[/tex] and [tex]\( 3x - ky = 0 \)[/tex] is [tex]\(\tan^{-1}\left(\frac{11}{2}\right)\)[/tex], follow these steps:
### Step 1: Find the slopes of the lines
- For the line [tex]\( 2x + y = 5 \)[/tex]:
[tex]\[
y = -2x + 5 \quad \Rightarrow \quad \text{slope } m_1 = -2
\][/tex]
- For the line [tex]\( 3x - ky = 0 \)[/tex]:
[tex]\[
ky = 3x \quad \Rightarrow \quad y = \frac{3}{k} x \quad \Rightarrow \quad \text{slope } m_2 = \frac{3}{k}
\][/tex]
### Step 2: Use the formula for the tangent of the angle between two lines
The angle [tex]\(\theta\)[/tex] between two lines with slopes [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] is given by:
[tex]\[
\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
\][/tex]
Given that:
[tex]\[
\tan(\theta) = \frac{11}{2}
\][/tex]
### Step 3: Substitute the slopes into the formula
[tex]\[
\tan\left(\theta\right) = \left| \frac{-2 - \frac{3}{k}}{1 + \left(-2 \cdot \frac{3}{k}\right)} \right|
\][/tex]
### Step 4: Simplify the expression
[tex]\[
\tan\left(\theta\right) = \left| \frac{ -2 - \frac{3}{k} }{ 1 - \frac{6}{k} } \right|
\][/tex]
### Step 5: Equate to [tex]\(\frac{11}{2}\)[/tex]
[tex]\[
\frac{11}{2} = \left| \frac{ -2k - 3 }{ k - 6 } \right|
\][/tex]
### Step 6: Solve the equation for [tex]\( k \)[/tex]
We set up two possible equations due to the absolute value:
1. [tex]\[
\frac{11}{2} = \frac{ -2k - 3 }{ k - 6 }
\][/tex]
2. [tex]\[
\frac{11}{2} = \frac{ 2k + 3 }{ 6 - k }
\][/tex]
#### Case 1:
[tex]\[
\frac{11}{2} = \frac{ -2k - 3 }{ k - 6 }
\][/tex]
Cross-multiplying gives:
[tex]\[
11(k - 6) = 2(-2k - 3)
\][/tex]
[tex]\[
11k - 66 = -4k - 6
\][/tex]
[tex]\[
11k + 4k = 66 - 6
\][/tex]
[tex]\[
15k = 60
\][/tex]
[tex]\[
k = 4
\][/tex]
#### Case 2:
[tex]\[
\frac{11}{2} = \frac{ 2k + 3 }{ 6 - k }
\][/tex]
Cross-multiplying gives:
[tex]\[
11(6 - k) = 2(2k + 3)
\][/tex]
[tex]\[
66 - 11k = 4k + 6
\][/tex]
[tex]\[
66 - 6 = 4k + 11k
\][/tex]
[tex]\[
60 = 15k
\][/tex]
[tex]\[
k = 4
\][/tex]
### Conclusion
Both cases yield [tex]\( k = 4 \)[/tex]. Therefore, the value of [tex]\( k \)[/tex] that makes the acute angle between the lines [tex]\( 2x + y = 5 \)[/tex] and [tex]\( 3x - ky = 0 \)[/tex] equal to [tex]\(\tan^{-1}\left(\frac{11}{2}\right)\)[/tex] is
[tex]\[
k = 4
\][/tex]
