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3D Geometry question

2024 · 1 Feb · Shift 1 · Q50
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  5. /2024 · 1 Feb · Shift 1 · Q50

3D Geometry question

2024 · 1 Feb · Shift 1 · Q50

JEE MainMathematics3D GeometryMCQ+4 / −1
If the shortest distance between the lines x−λ−2=y−21=z−11\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}−2x−λ​=1y−2​=1z−1​ and x−31=y−1−2=z−21\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}1x−3​​=−2y−1​=1z−2​ is 1 , then the sum of all possible values of λ\lambdaλ is :
  1. A
    0
  2. B
    232 \sqrt{3}23​
  3. C
    333 \sqrt{3}33​
  4. D
    −23-2 \sqrt{3}−23​
View written solutionFree

Correct answer: B

  1. Write the two lines in vector form

The lines are

x−λ−2=y−21=z−11\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}−2x−λ​=1y−2​=1z−1​

and

x−31=y−1−2=z−21.\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}.1x−3​​=−2y−1​=1z−2​.

So we can take:

  • A point on line L1L_1L1​ as A=(λ,2,1)A=(\lambda,2,1)A=(λ,2,1) and its direction vector as
d⃗1=(−2,1,1).\vec d_1=(-2,1,1).d1​=(−2,1,1).
  • A point on line L2L_2L2​ as B=(3,1,2)B=(\sqrt3,1,2)B=(3​,1,2) and its direction vector as
d⃗2=(1,−2,1).\vec d_2=(1,-2,1).d2​=(1,−2,1).
  1. Use formula for shortest distance between two skew lines

Shortest distance between lines L1L_1L1​ and L2L_2L2​ is

D=∣(AB→)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.D=\frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.D=∣d1​×d2​∣∣(AB)⋅(d1​×d2​)∣​.

Here,

AB→=B−A=(3−λ,−1,1).\overrightarrow{AB}=B-A=(\sqrt3-\lambda,-1,1).AB=B−A=(3​−λ,−1,1).
  1. Compute the cross product d⃗1×d⃗2\vec d_1\times \vec d_2d1​×d2​
d⃗1×d⃗2=∣i^j^k^−2111−21∣\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ -2 & 1 & 1\\ 1 & -2 & 1 \end{vmatrix}d1​×d2​=​i^−21​j^​1−2​k^11​​ =i^(1⋅1−1⋅(−2))−j^((−2)⋅1−1⋅1)+k^((−2)(−2)−1⋅1)=\hat i(1\cdot1-1\cdot(-2)) -\hat j((-2)\cdot1-1\cdot1) +\hat k((-2)(-2)-1\cdot1)=i^(1⋅1−1⋅(−2))−j^​((−2)⋅1−1⋅1)+k^((−2)(−2)−1⋅1) =3i^+3j^+3k^=(3,3,3).=3\hat i+3\hat j+3\hat k=(3,3,3).=3i^+3j^​+3k^=(3,3,3).

Hence,

∣d⃗1×d⃗2∣=32+32+32=33.|\vec d_1\times \vec d_2|=\sqrt{3^2+3^2+3^2}=3\sqrt3.∣d1​×d2​∣=32+32+32​=33​.
  1. Compute the scalar triple product
AB→⋅(d⃗1×d⃗2)=(3−λ,−1,1)⋅(3,3,3)\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2) =(\sqrt3-\lambda,-1,1)\cdot(3,3,3)AB⋅(d1​×d2​)=(3​−λ,−1,1)⋅(3,3,3) =3(3−λ)−3+3=3(3−λ).=3(\sqrt3-\lambda)-3+3=3(\sqrt3-\lambda).=3(3​−λ)−3+3=3(3​−λ).

So the shortest distance is

D=\frac{|3(\sqrt3-\lambda)|}{3\sqrt3}= rac{|\sqrt3-\lambda|}{\sqrt3}.

Given D=1D=1D=1, therefore

∣3−λ∣3=1.\frac{|\sqrt3-\lambda|}{\sqrt3}=1.3​∣3​−λ∣​=1.

So,

∣3−λ∣=3.|\sqrt3-\lambda|=\sqrt3.∣3​−λ∣=3​.

This gives two cases:

  • 3−λ=3  ⟹  λ=0\sqrt3-\lambda=\sqrt3 \implies \lambda=03​−λ=3​⟹λ=0
  • 3−λ=−3  ⟹  λ=23\sqrt3-\lambda=-\sqrt3 \implies \lambda=2\sqrt33​−λ=−3​⟹λ=23​
  1. Sum of all possible values of λ\lambdaλ
0+23=23.0+2\sqrt3=2\sqrt3.0+23​=23​.
  1. Compare with given options

Thus the correct option is

23\boxed{2\sqrt3}23​​

which is Option B.

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