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MathMedAdv 2014 • Paper 1

Find Parameter Lambda for Perpendicular Lines from Point to Given Lines

From a point P(λ,λ,λ)P(\lambda, \lambda, \lambda), perpendicular lines PQPQ and PRPR are drawn to the lines y=x,z=1y = x, z = 1 and y=−x,z=−1y = -x, z = -1 respectively. Given that ∠QPR\angle QPR is a right angle, find the potential value(s) of λ\lambda.

Options

A

2\sqrt{2}

B

11

C

−1-1

Correct
D

−2-\sqrt{2}

Topics & Concepts

Three Dimensional GeometryLine in 3DDirection Cosines and Ratios

Step-by-Step Solution

To find the potential value(s) of λ\lambda, we analyze the geometry of the given lines and the conditions on the point P(λ,λ,λ)P(\lambda, \lambda, \lambda).

Step 1: Find the foot of perpendicular QQ on Line 1 (L1L_1)

The equation of line L1L_1 is given by y=xy = x and z=1z = 1. In vector/parametric form, any point QQ on L1L_1 can be expressed as: Q=(t,t,1)for some t∈RQ = (t, t, 1) \quad \text{for some } t \in \mathbb{R}

The direction vector of L1L_1 is d⃗1=(1,1,0)\vec{d}_1 = (1, 1, 0).
The vector PQ⃗\vec{PQ} connecting P(λ,λ,λ)P(\lambda, \lambda, \lambda) to QQ is: PQ⃗=(t−λ)i^+(t−λ)j^+(1−λ)k^\vec{PQ} = (t - \lambda)\hat{i} + (t - \lambda)\hat{j} + (1 - \lambda)\hat{k}

Since PQ⊥L1PQ \perp L_1, their dot product must be zero: PQ⃗⋅d⃗1=0  ⟹  (t−λ)(1)+(t−λ)(1)+(1−λ)(0)=0\vec{PQ} \cdot \vec{d}_1 = 0 \implies (t - \lambda)(1) + (t - \lambda)(1) + (1 - \lambda)(0) = 0 2(t−λ)=0  ⟹  t=λ2(t - \lambda) = 0 \implies t = \lambda

Substituting t=λt = \lambda gives the coordinates of QQ as (λ,λ,1)(\lambda, \lambda, 1), and the vector PQ⃗\vec{PQ} becomes: PQ⃗=0i^+0j^+(1−λ)k^=(1−λ)k^\vec{PQ} = 0\hat{i} + 0\hat{j} + (1 - \lambda)\hat{k} = (1 - \lambda)\hat{k}

Step 2: Find the foot of perpendicular RR on Line 2 (L2L_2)

The equation of line L2L_2 is given by y=−xy = -x and z=−1z = -1. In vector/parametric form, any point RR on L2L_2 can be expressed as: R=(s,−s,−1)for some s∈RR = (s, -s, -1) \quad \text{for some } s \in \mathbb{R}

The direction vector of L2L_2 is d⃗2=(1,−1,0)\vec{d}_2 = (1, -1, 0).
The vector PR⃗\vec{PR} connecting P(λ,λ,λ)P(\lambda, \lambda, \lambda) to RR is: PR⃗=(s−λ)i^+(−s−λ)j^+(−1−λ)k^\vec{PR} = (s - \lambda)\hat{i} + (-s - \lambda)\hat{j} + (-1 - \lambda)\hat{k}

Since PR⊥L2PR \perp L_2, their dot product must be zero: PR⃗⋅d⃗2=0  ⟹  (s−λ)(1)+(−s−λ)(−1)+(−1−λ)(0)=0\vec{PR} \cdot \vec{d}_2 = 0 \implies (s - \lambda)(1) + (-s - \lambda)(-1) + (-1 - \lambda)(0) = 0 (s−λ)+(s+λ)=0  ⟹  2s=0  ⟹  s=0(s - \lambda) + (s + \lambda) = 0 \implies 2s = 0 \implies s = 0

Substituting s=0s = 0 gives the coordinates of RR as (0,0,−1)(0, 0, -1), and the vector PR⃗\vec{PR} becomes: PR⃗=−λi^−λj^−(1+λ)k^\vec{PR} = -\lambda\hat{i} - \lambda\hat{j} - (1 + \lambda)\hat{k}

Step 3: Apply the condition ∠QPR=90∘\angle QPR = 90^\circ

Since ∠QPR\angle QPR is a right angle, the vectors PQ⃗\vec{PQ} and PR⃗\vec{PR} must be non-zero and orthogonal to each other: PQ⃗⋅PR⃗=0\vec{PQ} \cdot \vec{PR} = 0

Substituting the vectors PQ⃗\vec{PQ} and PR⃗\vec{PR}: 0⋅(−λ)+0⋅(−λ)+(1−λ)(−(1+λ))=00 \cdot (-\lambda) + 0 \cdot (-\lambda) + (1 - \lambda)(-(1 + \lambda)) = 0 −(1−λ)(1+λ)=0-(1 - \lambda)(1 + \lambda) = 0 λ2−1=0  ⟹  λ=±1\lambda^2 - 1 = 0 \implies \lambda = \pm 1

Step 4: Validate the solutions

  • Case 1: λ=1\lambda = 1
    If λ=1\lambda = 1, then PQ⃗=(1−1)k^=0⃗\vec{PQ} = (1 - 1)\hat{k} = \vec{0}. This implies that the point PP coincides with QQ (i.e., PP lies on line L1L_1). Thus, no distinct line segment PQPQ exists, and the angle ∠QPR\angle QPR is undefined. Therefore, λ=1\lambda = 1 is discarded.

  • Case 2: λ=−1\lambda = -1
    If λ=−1\lambda = -1, we get: PQ⃗=2k^≠0⃗\vec{PQ} = 2\hat{k} \neq \vec{0} PR⃗=i^+j^≠0⃗\vec{PR} = \hat{i} + \hat{j} \neq \vec{0} Here, PQ⃗⋅PR⃗=0(1)+0(1)+2(0)=0\vec{PQ} \cdot \vec{PR} = 0(1) + 0(1) + 2(0) = 0. Points P,Q,RP, Q, R form a non-degenerate right-angled triangle with ∠QPR=90∘\angle QPR = 90^\circ.

Thus, the only valid value of λ\lambda is −1-1.

Correct Answer: (C) −1-1

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