The plane lx + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. The equation of the plane in its new position is lx + my ± z\(\sqrt{l^{2}+m^{2}}\) ⋅ λ = 0, then λ equals

Concept Used:

Let A1 x + B1 y + C1z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0  be the equation of two planes aligned to each other at an angle θ

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

The plane lx + my = 0 . . .(i) has been rotated about its line of intersection with the plane z = 0; hence it will pass through the intersection of the plane lx+ my = 0 and z = 0.
Let the equation of the plane after rotation be lx + my + λz = 0. . .(ii)
Then angle between plane and line is α

\( \therefore \cos \alpha=\frac{l\cdot l+\mathrm{m} \cdot \mathrm{m}+λ \cdot 0}{\sqrt{\left(l^2+\mathrm{m}^2+\mathrm{λ}^2\right)} \sqrt{\left(l^2+\mathrm{m}^2\right)}}\)

\( \Rightarrow \cos \alpha=\frac{\sqrt{\left(l^2+\mathrm{m}^2\right)}}{\sqrt{l^2+\mathrm{m}^2+λ^2}}\)

\( \Rightarrow \left(l^2+\mathrm{m}^2+λ^2\right) \cos ^2 \alpha=l^2+\mathrm{m}^2 \\ \)

\( \Rightarrow λ^2 \cos ^2 \alpha=\left(l^2+\mathrm{m}^2\right)\left(1-\cos ^2 \alpha \right), \)

\( \Rightarrow λ^2=\left(l^2+\mathrm{m}^2\right) \tan ^2 \alpha\)

\( \Rightarrow λ=\pm \sqrt{l^2+\mathrm{m}^2} \tan \alpha\)

Putting this value of λ in (ii), the required plane is given by,

\( l \mathrm{x}+\mathrm{my} \pm \mathrm{z} \sqrt{l^2+\mathrm{m}^2} \tan \alpha=0\)