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3D Geometry question

2024 · 29 Jan · Shift 2 · Q51
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  5. /2024 · 29 Jan · Shift 2 · Q51

3D Geometry question

2024 · 29 Jan · Shift 2 · Q51

JEE MainMathematics3D GeometryNumerical+4 / −1
Let O be the origin, and M and N\mathrm{N}N be the points on the lines x−54=y−41=z−53\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}4x−5​=1y−4​=3z−5​ and x+812=y+25=z+119\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}12x+8​=5y+2​=9z+11​ respectively such that MN\mathrm{MN}MN is the shortest distance between the given lines. Then OM→⋅ON→\overrightarrow{O M} \cdot \overrightarrow{O N}OM⋅ON is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 9

  1. Write the two lines in vector form

The given lines are

x−54=y−41=z−53\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}4x−5​=1y−4​=3z−5​

and

x+812=y+25=z+119.\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}.12x+8​=5y+2​=9z+11​.

So their parametric forms are:

  • Line L1L_1L1​: (x,y,z)=(5,4,5)+λ(4,1,3)(x,y,z)=(5,4,5)+\lambda(4,1,3)(x,y,z)=(5,4,5)+λ(4,1,3)
  • Line L2L_2L2​: (x,y,z)=(−8,−2,−11)+μ(12,5,9)(x,y,z)=(-8,-2,-11)+\mu(12,5,9)(x,y,z)=(−8,−2,−11)+μ(12,5,9)

Let

M=(5,4,5)+λ(4,1,3),N=(−8,−2,−11)+μ(12,5,9).M=(5,4,5)+\lambda(4,1,3), \qquad N=(-8,-2,-11)+\mu(12,5,9).M=(5,4,5)+λ(4,1,3),N=(−8,−2,−11)+μ(12,5,9).
  1. Use the condition for shortest distance

If MNMNMN is the shortest distance between two skew lines, then the vector MN→\overrightarrow{MN}MN is perpendicular to both direction vectors.

Direction vectors are:

d⃗1=(4,1,3),d⃗2=(12,5,9).\vec d_1=(4,1,3), \qquad \vec d_2=(12,5,9).d1​=(4,1,3),d2​=(12,5,9).

Now,

M=(5+4λ, 4+λ, 5+3λ),M=(5+4\lambda,\ 4+\lambda,\ 5+3\lambda),M=(5+4λ, 4+λ, 5+3λ), N=(−8+12μ, −2+5μ, −11+9μ).N=(-8+12\mu,\ -2+5\mu,\ -11+9\mu).N=(−8+12μ, −2+5μ, −11+9μ).

Hence

MN→=N−M\overrightarrow{MN}=N-MMN=N−M

so

MN→=(−13+12μ−4λ, −6+5μ−λ, −16+9μ−3λ).\overrightarrow{MN}=(-13+12\mu-4\lambda,\ -6+5\mu-\lambda,\ -16+9\mu-3\lambda).MN=(−13+12μ−4λ, −6+5μ−λ, −16+9μ−3λ).

Since MN→⊥d⃗1\overrightarrow{MN}\perp \vec d_1MN⊥d1​ and MN→⊥d⃗2\overrightarrow{MN}\perp \vec d_2MN⊥d2​,

MN→⋅(4,1,3)=0\overrightarrow{MN}\cdot (4,1,3)=0MN⋅(4,1,3)=0

and

MN→⋅(12,5,9)=0.\overrightarrow{MN}\cdot (12,5,9)=0.MN⋅(12,5,9)=0.
  1. Form the equations

First,

(−13+12μ−4λ)4+(−6+5μ−λ)+(−16+9μ−3λ)3=0.(-13+12\mu-4\lambda)4+(-6+5\mu-\lambda)+(-16+9\mu-3\lambda)3=0.(−13+12μ−4λ)4+(−6+5μ−λ)+(−16+9μ−3λ)3=0.

Expanding:

−52+48μ−16λ−6+5μ−λ−48+27μ−9λ=0-52+48\mu-16\lambda-6+5\mu-\lambda-48+27\mu-9\lambda=0−52+48μ−16λ−6+5μ−λ−48+27μ−9λ=0 80μ−26λ−106=080\mu-26\lambda-106=080μ−26λ−106=0 40μ−13λ=53.(1)40\mu-13\lambda=53. \quad (1)40μ−13λ=53.(1)

Second,

(−13+12μ−4λ)12+(−6+5μ−λ)5+(−16+9μ−3λ)9=0.(-13+12\mu-4\lambda)12+(-6+5\mu-\lambda)5+(-16+9\mu-3\lambda)9=0.(−13+12μ−4λ)12+(−6+5μ−λ)5+(−16+9μ−3λ)9=0.

Expanding:

−156+144μ−48λ−30+25μ−5λ−144+81μ−27λ=0-156+144\mu-48\lambda-30+25\mu-5\lambda-144+81\mu-27\lambda=0−156+144μ−48λ−30+25μ−5λ−144+81μ−27λ=0 250μ−80λ−330=0250\mu-80\lambda-330=0250μ−80λ−330=0 25μ−8λ=33.(2)25\mu-8\lambda=33. \quad (2)25μ−8λ=33.(2)
  1. Solve for λ\lambdaλ and μ\muμ

From (1) and (2):

40μ−13λ=53,40\mu-13\lambda=53,40μ−13λ=53, 25μ−8λ=33.25\mu-8\lambda=33.25μ−8λ=33.

Multiply the first by 888 and the second by 131313:

320μ−104λ=424,320\mu-104\lambda=424,320μ−104λ=424, 325μ−104λ=429.325\mu-104\lambda=429.325μ−104λ=429.

Subtracting,

5μ=5⇒μ=1.5\mu=5 \Rightarrow \mu=1.5μ=5⇒μ=1.

Substitute into (2):

25(1)−8λ=3325(1)-8\lambda=3325(1)−8λ=33 −8λ=8-8\lambda=8−8λ=8 λ=−1.\lambda=-1.λ=−1.
  1. Find points MMM and NNN

For λ=−1\lambda=-1λ=−1,

M=(5,4,5)+(−1)(4,1,3)=(1,3,2).M=(5,4,5)+(-1)(4,1,3)=(1,3,2).M=(5,4,5)+(−1)(4,1,3)=(1,3,2).

For μ=1\mu=1μ=1,

N=(−8,−2,−11)+(12,5,9)=(4,3,−2).N=(-8,-2,-11)+(12,5,9)=(4,3,-2).N=(−8,−2,−11)+(12,5,9)=(4,3,−2).

Thus,

OM→=(1,3,2),ON→=(4,3,−2).\overrightarrow{OM}=(1,3,2), \qquad \overrightarrow{ON}=(4,3,-2).OM=(1,3,2),ON=(4,3,−2).
  1. Compute the dot product
OM→⋅ON→=(1)(4)+(3)(3)+(2)(−2)\overrightarrow{OM}\cdot \overrightarrow{ON}=(1)(4)+(3)(3)+(2)(-2)OM⋅ON=(1)(4)+(3)(3)+(2)(−2) =4+9−4=9.=4+9-4=9.=4+9−4=9.
  1. Final answer
9\boxed{9}9​

This matches the stored correct answer.

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