If the lines $\dfrac{x}{3} + \dfrac{y}{4} = 7$ and 3x + ky = 11 are perpendicular to each other, find the value of k.

Given, $\dfrac{x}{3} + \dfrac{y}{4} = 7$ and 3x + ky = 11.

Converting $\dfrac{x}{3} + \dfrac{y}{4} = 7$ in the form of y = mx + c.

$\Rightarrow \dfrac{4x + 3y}{12} = 7 \\[1em] \Rightarrow 4x + 3y = 84 \\[1em] \Rightarrow 3y = -4x + 84 \\[1em] \Rightarrow y = -\dfrac{4}{3}x + 28.$

Comparing above equation with y = mx + c we get slope (m₁),

$m_1 = -\dfrac{4}{3}$.

Converting 3x + ky = 11 in the form of y = mx + c.

$\Rightarrow 3x + ky = 11 \\[1em] \Rightarrow ky = -3x + 11 \\[1em] \Rightarrow y = -\dfrac{3}{k}x + \dfrac{11}{k}.$

Comparing above equation with y = mx + c we get slope,

$m_2 = -\dfrac{3}{k}$.

Given two lines are perpendicular,

∴ m₁ × m₂ = -1.

$\Rightarrow -\dfrac{4}{3} \times -\dfrac{3}{k} = -1 \\[1em] \Rightarrow \dfrac{4}{k} = -1 \\[1em] \Rightarrow k = -4.$

Hence, the value of k is -4.