Solveeit Logo
Login

Question

Question: If two lines $L_1$ and $L_2$ in space, are defined by $L_1 = \{x = \sqrt{\lambda}y + (\sqrt{\lambda}...

If two lines L1L_1 and L2L_2 in space, are defined by L1={x=λy+(λ−1),z=(λ−1)y+λ}L_1 = \{x = \sqrt{\lambda}y + (\sqrt{\lambda}-1), z = (\sqrt{\lambda}-1)y + \sqrt{\lambda}\} and L2={x=μy+(1−μ),z=(1−μ)y+μ}L_2 = \{x = \sqrt{\mu}y + (1 - \sqrt{\mu}), z = (1 - \sqrt{\mu})y + \sqrt{\mu}\} then L1L_1 is perpendicular to L2L_2, for all non-negative reals λ\lambda and μ\mu, such that:

A

λ+μ=1\sqrt{\lambda} + \sqrt{\mu} = 1

B

λ≠μ\lambda \neq \mu

C

λ+μ=0\lambda + \mu = 0

D

λ=μ\lambda = \mu

Answer

(c)

Explanation

Solution

The direction vectors of the lines L1L_1 and L2L_2 are v1⃗=⟨λ,1,λ−1⟩\vec{v_1} = \langle \sqrt{\lambda}, 1, \sqrt{\lambda}-1 \rangle and v2⃗=⟨μ,1,1−μ⟩\vec{v_2} = \langle \sqrt{\mu}, 1, 1 - \sqrt{\mu} \rangle, respectively. For the lines to be perpendicular, their direction vectors must be orthogonal, meaning their dot product is zero. The dot product v1⃗⋅v2⃗=λ+μ\vec{v_1} \cdot \vec{v_2} = \sqrt{\lambda} + \sqrt{\mu}. Thus, perpendicularity requires λ+μ=0\sqrt{\lambda} + \sqrt{\mu} = 0. Since λ,μ≥0\lambda, \mu \ge 0, this condition is met if and only if λ=0\lambda=0 and μ=0\mu=0. Among the given options, only λ+μ=0\lambda + \mu = 0 implies λ=0\lambda=0 and μ=0\mu=0 for non-negative reals.