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3D Geometry question

2025 · 23 Jan · Shift 2 · Q31
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  5. /2025 · 23 Jan · Shift 2 · Q31

3D Geometry question

2025 · 23 Jan · Shift 2 · Q31

JEE MainMathematics3D GeometryMCQ+4 / −1
If the square of the shortest distance between the lines x−21=y−12=z+3−3\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}1x−2​=2y−1​=−3z+3​ and x+12=y+34=z+5−5\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}2x+1​=4y+3​=−5z+5​ is mn\frac{m}{n}nm​, where mmm, nnn are coprime numbers, then m+nm+nm+n is equal to :
  1. A
    14
  2. B
    6
  3. C
    21
  4. D
    9
View written solutionFree

Correct answer: D

  1. Write the lines in vector form

The given lines are

x−21=y−12=z+3−3\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}1x−2​=2y−1​=−3z+3​

and

x+12=y+34=z+5−5.\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}.2x+1​=4y+3​=−5z+5​.

So their parametric/vector forms are:

  • Line L1L_1L1​ passes through A=(2,1,−3)A=(2,1,-3)A=(2,1,−3) with direction vector d⃗1=(1,2,−3).\vec d_1=(1,2,-3).d1​=(1,2,−3).

  • Line L2L_2L2​ passes through B=(−1,−3,−5)B=(-1,-3,-5)B=(−1,−3,−5) with direction vector d⃗2=(2,4,−5).\vec d_2=(2,4,-5).d2​=(2,4,−5).

  1. Formula for shortest distance between two skew lines

For lines through points A,BA,BA,B with direction vectors d⃗1,d⃗2\vec d_1, \vec d_2d1​,d2​, the shortest distance 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=(−1−2, −3−1, −5−(−3))=(−3,−4,−2).\overrightarrow{AB}=B-A=(-1-2,\,-3-1,\,-5-(-3))=(-3,-4,-2).AB=B−A=(−1−2,−3−1,−5−(−3))=(−3,−4,−2).
  1. Compute the cross product
d⃗1×d⃗2=∣i^j^k^12−324−5∣\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}d1​×d2​=​i^12​j^​24​k^−3−5​​ =i^ (2⋅(−5)−(−3)⋅4)−j^ (1⋅(−5)−(−3)⋅2)+k^ (1⋅4−2⋅2)=\hat i\,(2\cdot(-5)-(-3)\cdot 4) -\hat j\,(1\cdot(-5)-(-3)\cdot 2) +\hat k\,(1\cdot 4-2\cdot 2)=i^(2⋅(−5)−(−3)⋅4)−j^​(1⋅(−5)−(−3)⋅2)+k^(1⋅4−2⋅2) =i^(−10+12)−j^(−5+6)+k^(4−4)=(2,−1,0).=\hat i(-10+12)-\hat j(-5+6)+\hat k(4-4) =(2,-1,0).=i^(−10+12)−j^​(−5+6)+k^(4−4)=(2,−1,0).

Thus,

∣d⃗1×d⃗2∣=22+(−1)2+02=5.|\vec d_1\times \vec d_2|=\sqrt{2^2+(-1)^2+0^2}=\sqrt{5}.∣d1​×d2​∣=22+(−1)2+02​=5​.
  1. Compute the scalar triple product
AB→⋅(d⃗1×d⃗2)=(−3,−4,−2)⋅(2,−1,0)\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)=(-3,-4,-2)\cdot(2,-1,0)AB⋅(d1​×d2​)=(−3,−4,−2)⋅(2,−1,0) =−3⋅2+(−4)(−1)+(−2)⋅0=−6+4=−2.=-3\cdot 2+(-4)(-1)+(-2)\cdot 0=-6+4=-2.=−3⋅2+(−4)(−1)+(−2)⋅0=−6+4=−2.

Hence,

∣AB→⋅(d⃗1×d⃗2)∣=2.|\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)|=2.∣AB⋅(d1​×d2​)∣=2.

So the shortest distance is

D=25.D=\frac{2}{\sqrt{5}}.D=5​2​.
  1. Square of the shortest distance
D2=(25)2=45.D^2=\left(\frac{2}{\sqrt{5}}\right)^2=\frac{4}{5}.D2=(5​2​)2=54​.

Thus,

mn=45\frac{m}{n}=\frac{4}{5}nm​=54​

with coprime m=4m=4m=4, n=5n=5n=5.

Therefore,

m+n=4+5=9.m+n=4+5=9.m+n=4+5=9.
  1. Compare with options

The correct option is: D: 9\boxed{\text{D: }9}D: 9​

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