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3D Geometry question

2014 · Shift 1 · Q29
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  5. /2014 · Shift 1 · Q29

3D Geometry question

2014 · Shift 1 · Q29

JEE AdvancedMathematics3D GeometryMultiple correct+3 / −1
From a point P(λ,λ,λ),P\left( {\lambda ,\lambda ,\lambda } \right),P(λ,λ,λ), perpendicular PQPQPQ and PRPRPR are drawn respectively on the lines y=x,z=1y=x, z=1y=x,z=1 and y=−x,z=−1.y=-x, z=-1.y=−x,z=−1. If PPP is such that ∠QPR\angle QPR∠QPR is a right angle, then the possible value(s) of λ\lambdaλ is/(are)
  1. A
    2\sqrt 22​
  2. B
    111
  3. C
    −1-1−1
  4. D
    −2-\sqrt 2−2​
View written solutionFree

Correct answer: B, C

  1. Interpret the two lines in 3D

The lines are given as:

  • L1:y=x, z=1L_1: y=x,\ z=1L1​:y=x, z=1
  • L2:y=−x, z=−1L_2: y=-x,\ z=-1L2​:y=−x, z=−1

We parametrize them as:

L1:(t,t,1),direction vector d⃗1=(1,1,0)L_1: (t,t,1), \quad \text{direction vector } \vec d_1=(1,1,0)L1​:(t,t,1),direction vector d1​=(1,1,0)

L2:(s,−s,−1),direction vector d⃗2=(1,−1,0)L_2: (s,-s,-1), \quad \text{direction vector } \vec d_2=(1,-1,0)L2​:(s,−s,−1),direction vector d2​=(1,−1,0)

The point is

P=(λ,λ,λ).P=(\lambda,\lambda,\lambda).P=(λ,λ,λ).

  1. Find the foot of perpendicular from PPP to L1L_1L1​

Let Q=(t,t,1)Q=(t,t,1)Q=(t,t,1) be the foot on L1L_1L1​. Then

PQ⃗=Q−P=(t−λ, t−λ, 1−λ).\vec{PQ}=Q-P=(t-\lambda,\ t-\lambda,\ 1-\lambda).PQ​=Q−P=(t−λ, t−λ, 1−λ).

Since PQ⊥L1PQ \perp L_1PQ⊥L1​, we must have

PQ⃗⋅d⃗1=0.\vec{PQ}\cdot \vec d_1=0.PQ​⋅d1​=0.

So,

(t−λ, t−λ, 1−λ)⋅(1,1,0)=0(t-\lambda,\ t-\lambda,\ 1-\lambda)\cdot (1,1,0)=0(t−λ, t−λ, 1−λ)⋅(1,1,0)=0

(t−λ)+(t−λ)=0(t-\lambda)+(t-\lambda)=0(t−λ)+(t−λ)=0

2(t−λ)=0⇒t=λ.2(t-\lambda)=0 \Rightarrow t=\lambda.2(t−λ)=0⇒t=λ.

Hence,

Q=(λ,λ,1).Q=(\lambda,\lambda,1).Q=(λ,λ,1).

Therefore,

PQ⃗=(0,0,1−λ).\vec{PQ}=(0,0,1-\lambda).PQ​=(0,0,1−λ).

  1. Find the foot of perpendicular from PPP to L2L_2L2​

Let R=(s,−s,−1)R=(s,-s,-1)R=(s,−s,−1) be the foot on L2L_2L2​. Then

PR⃗=R−P=(s−λ, −s−λ, −1−λ).\vec{PR}=R-P=(s-\lambda,\ -s-\lambda,\ -1-\lambda).PR=R−P=(s−λ, −s−λ, −1−λ).

Since PR⊥L2PR \perp L_2PR⊥L2​, we must have

PR⃗⋅d⃗2=0.\vec{PR}\cdot \vec d_2=0.PR⋅d2​=0.

So,

(s−λ, −s−λ, −1−λ)⋅(1,−1,0)=0(s-\lambda,\ -s-\lambda,\ -1-\lambda)\cdot (1,-1,0)=0(s−λ, −s−λ, −1−λ)⋅(1,−1,0)=0

(s−λ)+(s+λ)=0(s-\lambda)+ (s+\lambda)=0(s−λ)+(s+λ)=0

2s=0⇒s=0.2s=0 \Rightarrow s=0.2s=0⇒s=0.

Hence,

R=(0,0,−1).R=(0,0,-1).R=(0,0,−1).

Therefore,

PR⃗=(−λ,−λ,−1−λ).\vec{PR}=(-\lambda,-\lambda,-1-\lambda).PR=(−λ,−λ,−1−λ).

  1. Use the condition ∠QPR=90∘\angle QPR=90^\circ∠QPR=90∘

Since the angle between PQPQPQ and PRPRPR is a right angle,

PQ⃗⋅PR⃗=0.\vec{PQ}\cdot \vec{PR}=0.PQ​⋅PR=0.

Now,

PQ⃗=(0,0,1−λ),PR⃗=(−λ,−λ,−1−λ).\vec{PQ}=(0,0,1-\lambda), \qquad \vec{PR}=(-\lambda,-\lambda,-1-\lambda).PQ​=(0,0,1−λ),PR=(−λ,−λ,−1−λ).

Thus,

PQ⃗⋅PR⃗=0+0+(1−λ)(−1−λ)=0.\vec{PQ}\cdot \vec{PR}=0+0+(1-\lambda)(-1-\lambda)=0.PQ​⋅PR=0+0+(1−λ)(−1−λ)=0.

So,

(1−λ)(−1−λ)=0.(1-\lambda)(-1-\lambda)=0.(1−λ)(−1−λ)=0.

This gives

λ=1orλ=−1.\lambda=1 \quad \text{or} \quad \lambda=-1.λ=1orλ=−1.

  1. Check the options

Options are:

  • A: 2\sqrt22​
  • B: 111
  • C: −1-1−1
  • D: −2-\sqrt2−2​

Hence the correct options are

1 and −1\boxed{1 \text{ and } -1}1 and −1​

So the correct choices are B and C.

  1. Comparison with stored correct answer

Stored correct answer: C

But our derived answer is B, C.

Therefore, the stored answer is incomplete/incorrect.

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