Let L1 and L2 be the following straight lines. L1: (x-1/1)=(y/-1)=(z-1/3) and L2: (x-1/-3)-(y/-1)=(z-1/1) . Suppose thestraight line L: (x-α/1)=(y-1/m)=(z-γ/-2) lies in the plane containing L1 and L2, and passes through thepoint of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?

Question

Let L1 and L2 be the following straight lines. L1: (x-1/1)=(y/-1)=(z-1/3) and L2: (x-1/-3)-(y/-1)=(z-1/1) . Suppose thestraight line L: (x-α/1)=(y-1/m)=(z-&gamma;/-2) lies in the plane containing L1 and L2, and passes through thepoint of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE? (A) α-&gamma;=3 (B) l+m=2 (C) α-&gamma;=1 (D) l+m=0

Tardigrade Admin · Accepted Answer

Clearly we can observe from the equations that

L_{1}

and

L_{2}

pass through

(1, 0, 1)

Vectors in the directions of

L_{1}

and

L_{2}

are

(\hat{i} - \hat{j} + 3 \hat{k})

and

(- 3 \hat{i} - \hat{j} + \hat{k})

respectively which areequimodular and inclined at an acute angle, so their angle bisector will be along their resultant, which is

(- 2 \hat{i} - 2 \hat{j} + 4 \hat{k})

. So equation of angle acute angle bisector will be

\frac{x - 1}{1} = \frac{y}{1} = \frac{z - 1}{- 2} (

Clearly

1 = m = 1)

∵

Point

(α, 1, γ)

lies on this line, then

α = 2

and

γ = - 1

$α - γ = 3$

$l + m = 2$

$α - γ = 1$

$l + m = 0$

α−γ=3

l+m=2

α−γ=1

l+m=0

$α - γ = 3$

$l + m = 2$

$α - γ = 1$

$l + m = 0$