Determine the value of λ if planes 2x + 4y – 4z = 6 and λx + 3y + 9 = 0 make an angle of  \({\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)\).

CONCEPT:

• Let A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂ z + D₂ = 0 are the equations of two planes aligned at an angle θ where A₁, B₁, C₁ and A₂, B₂, C₂ are the direction ratios of the normal to the planes, then the cosine of the angle between the two planes is given by:

\(cos\theta = \left| {\frac{{{A_1}{A_2} + {B_1}{B_2} + {C_1}{C_2}}}{{\sqrt {A_1^2 + B_1^2 + C_1^2} \sqrt {A_2^2 + B_2^2 + C_2^2} }}} \right|\)

CALCULATIONS:

Given planes are 2x + 4y – 4z = 6 and λx + 3y + 9 = 0,

Putting the value in equations –

\(\frac{1}{{\sqrt 2 }} = \left| {\frac{{2.\lambda + 4.3 - 4.\left( 0 \right)}}{{\sqrt {{2^2} + {{(4)}^2} + {{\left( { - 4} \right)}^2}} \sqrt {{\lambda ^2} + {3^2}} }}} \right| \Rightarrow \frac{1}{{\sqrt 2 }} = \left| {\frac{{2.\lambda + 12}}{{6.\sqrt {{\lambda ^2} + {3^2}} }}} \right|\)

∴ 7λ² -24λ + 9 = 0