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3D Geometry question

2024 · 8 Apr · Shift 2 · Q32
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  5. /2024 · 8 Apr · Shift 2 · Q32

3D Geometry question

2024 · 8 Apr · Shift 2 · Q32

JEE MainMathematics3D GeometryMCQ+4 / −1
If the shortest distance between the lines x−λ2=y−43=z−34\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}2x−λ​=3y−4​=4z−3​ and x−24=y−46=z−78\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}4x−2​=6y−4​=8z−7​ is 1329\frac{13}{\sqrt{29}}29​13​, then a value of λ\lambdaλ is :
  1. A
    1325\frac{13}{25}2513​
  2. B
    1
  3. C
    −-− 1
  4. D
    −1325-\frac{13}{25}−2513​
View written solutionFree

Correct answer: B

  1. Write the lines in vector form

The given lines are

x−λ2=y−43=z−34\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}2x−λ​=3y−4​=4z−3​

and

x−24=y−46=z−78.\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}.4x−2​=6y−4​=8z−7​.

So their parametric/vector forms are:

  • Line L1L_1L1​ passes through A(λ,4,3)A(\lambda,4,3)A(λ,4,3) and has direction vector
d⃗1=⟨2,3,4⟩.\vec d_1=\langle 2,3,4\rangle.d1​=⟨2,3,4⟩.
  • Line L2L_2L2​ passes through B(2,4,7)B(2,4,7)B(2,4,7) and has direction vector
d⃗2=⟨4,6,8⟩=2⟨2,3,4⟩.\vec d_2=\langle 4,6,8\rangle = 2\langle 2,3,4\rangle.d2​=⟨4,6,8⟩=2⟨2,3,4⟩.

Hence the lines are parallel.

  1. Formula for shortest distance between parallel lines

For two parallel lines with common direction vector d⃗\vec dd, shortest distance is

D=∣AB→×d⃗∣∣d⃗∣,D=\frac{|\overrightarrow{AB}\times \vec d|}{|\vec d|},D=∣d∣∣AB×d∣​,

where AAA is a point on the first line and BBB a point on the second.

Take

A=(λ,4,3),B=(2,4,7).A=(\lambda,4,3), \quad B=(2,4,7).A=(λ,4,3),B=(2,4,7).

Then

AB→=B−A=(2−λ,0,4).\overrightarrow{AB}=B-A=(2-\lambda,0,4).AB=B−A=(2−λ,0,4).

Also,

d⃗=(2,3,4).\vec d=(2,3,4).d=(2,3,4).
  1. Compute the cross product
AB→×d⃗=∣i^j^k^2−λ04234∣\overrightarrow{AB}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2-\lambda & 0 & 4\\ 2 & 3 & 4 \end{vmatrix}AB×d=​i^2−λ2​j^​03​k^44​​ =i^(0⋅4−4⋅3)−j^((2−λ)⋅4−4⋅2)+k^((2−λ)⋅3−0⋅2)=\hat i(0\cdot 4-4\cdot 3)-\hat j((2-\lambda)\cdot 4-4\cdot 2)+\hat k((2-\lambda)\cdot 3-0\cdot 2)=i^(0⋅4−4⋅3)−j^​((2−λ)⋅4−4⋅2)+k^((2−λ)⋅3−0⋅2) =−12i^−(8−4λ−8)j^+(6−3λ)k^=-12\hat i-\big(8-4\lambda-8\big)\hat j+(6-3\lambda)\hat k=−12i^−(8−4λ−8)j^​+(6−3λ)k^ =−12i^+4λj^+(6−3λ)k^.=-12\hat i+4\lambda\hat j+(6-3\lambda)\hat k.=−12i^+4λj^​+(6−3λ)k^.

So,

∣AB→×d⃗∣=(−12)2+(4λ)2+(6−3λ)2.|\overrightarrow{AB}\times \vec d|= \sqrt{(-12)^2+(4\lambda)^2+(6-3\lambda)^2}.∣AB×d∣=(−12)2+(4λ)2+(6−3λ)2​.

That is,

=144+16λ2+36−36λ+9λ2=\sqrt{144+16\lambda^2+36-36\lambda+9\lambda^2}=144+16λ2+36−36λ+9λ2​ =25λ2−36λ+180.=\sqrt{25\lambda^2-36\lambda+180}.=25λ2−36λ+180​.

Also,

∣d⃗∣=22+32+42=29.|\vec d|=\sqrt{2^2+3^2+4^2}=\sqrt{29}.∣d∣=22+32+42​=29​.

Thus shortest distance is

D=25λ2−36λ+18029.D=\frac{\sqrt{25\lambda^2-36\lambda+180}}{\sqrt{29}}.D=29​25λ2−36λ+180​​.

Given

D=1329.D=\frac{13}{\sqrt{29}}.D=29​13​.

So,

25λ2−36λ+180=13.\sqrt{25\lambda^2-36\lambda+180}=13.25λ2−36λ+180​=13.

Squaring,

25λ2−36λ+180=169.25\lambda^2-36\lambda+180=169.25λ2−36λ+180=169. 25λ2−36λ+11=0.25\lambda^2-36\lambda+11=0.25λ2−36λ+11=0.
  1. Solve the quadratic
25λ2−36λ+11=0.25\lambda^2-36\lambda+11=0.25λ2−36λ+11=0.

Using factorization,

25λ2−25λ−11λ+11=025\lambda^2-25\lambda-11\lambda+11=025λ2−25λ−11λ+11=0 25λ(λ−1)−11(λ−1)=025\lambda(\lambda-1)-11(\lambda-1)=025λ(λ−1)−11(λ−1)=0 (λ−1)(25λ−11)=0.(\lambda-1)(25\lambda-11)=0.(λ−1)(25λ−11)=0.

Hence,

λ=1orλ=1125.\lambda=1 \quad \text{or} \quad \lambda=\frac{11}{25}.λ=1orλ=2511​.
  1. Match with options

Given options are:

  • A: 1325\frac{13}{25}2513​
  • B: 111
  • C: −1-1−1
  • D: −1325-\frac{13}{25}−2513​

Only λ=1\lambda=1λ=1 is present in the options.

Therefore the correct option is

1.\boxed{1}.1​.
  1. Compare with stored answer

Stored correct answer: B

Our derived answer is B (λ=1\lambda=1λ=1), so it agrees.

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