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3D Geometry question

2025 · 8 Apr · Shift 2 · Q26
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  5. /2025 · 8 Apr · Shift 2 · Q26

3D Geometry question

2025 · 8 Apr · Shift 2 · Q26

JEE MainMathematics3D GeometryMCQ+4 / −1
Let the values of λ\lambdaλ for which the shortest distance between the lines x−12=y−23=z−34\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}2x−1​=3y−2​=4z−3​ and x−λ3=y−44=z−55\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}3x−λ​=4y−4​=5z−5​ is 16\frac{1}{\sqrt{6}}6​1​ be λ1\lambda_1λ1​ and λ2\lambda_2λ2​. Then the radius of the circle passing through the points (0,0),(λ1,λ2)(0, 0), (\lambda_1, \lambda_2)(0,0),(λ1​,λ2​) and (λ2,λ1)(\lambda_2, \lambda_1)(λ2​,λ1​) is
  1. A
    333
  2. B
    523\frac{5\sqrt{2}}{3}352​​
  3. C
    23\frac{\sqrt{2}}{3}32​​
  4. D
    444
View written solutionFree

Correct answer: B

  1. Write the two lines in vector form

The lines are

L1: x−12=y−23=z−34L_1:\ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}L1​: 2x−1​=3y−2​=4z−3​

and

L2: x−λ3=y−44=z−55.L_2:\ \frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}.L2​: 3x−λ​=4y−4​=5z−5​.

So,

  • A point on L1L_1L1​ is A=(1,2,3)A=(1,2,3)A=(1,2,3) and its direction vector is
d⃗1=(2,3,4).\vec d_1=(2,3,4).d1​=(2,3,4).
  • A point on L2L_2L2​ is B=(λ,4,5)B=(\lambda,4,5)B=(λ,4,5) and its direction vector is
d⃗2=(3,4,5).\vec d_2=(3,4,5).d2​=(3,4,5).
  1. Formula for shortest distance between two skew lines

The shortest distance between lines

r⃗=a⃗+td⃗1,r⃗=b⃗+sd⃗2\vec r=\vec a+t\vec d_1,\qquad \vec r=\vec b+s\vec d_2r=a+td1​,r=b+sd2​

is

D=∣(b⃗−a⃗)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.D=\frac{|(\vec b-\vec a)\cdot (\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.D=∣d1​×d2​∣∣(b−a)⋅(d1​×d2​)∣​.

Here,

AB⃗=B−A=(λ−1,2,2).\vec{AB}=B-A=(\lambda-1,2,2).AB=B−A=(λ−1,2,2).

Now compute the cross product:

d⃗1×d⃗2=∣i^j^k^234345∣=i^(15−16)−j^(10−12)+k^(8−9).\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} =\hat i(15-16)-\hat j(10-12)+\hat k(8-9).d1​×d2​=​i^23​j^​34​k^45​​=i^(15−16)−j^​(10−12)+k^(8−9).

Thus,

d⃗1×d⃗2=(−1,2,−1).\vec d_1\times \vec d_2=(-1,2,-1).d1​×d2​=(−1,2,−1).

Its magnitude is

∣d⃗1×d⃗2∣=(−1)2+22+(−1)2=6.|\vec d_1\times \vec d_2|=\sqrt{(-1)^2+2^2+(-1)^2}=\sqrt{6}.∣d1​×d2​∣=(−1)2+22+(−1)2​=6​.
  1. Use the given shortest distance

Given shortest distance is

16.\frac{1}{\sqrt 6}.6​1​.

So,

∣AB⃗⋅(d⃗1×d⃗2)∣6=16.\frac{|\vec{AB}\cdot(\vec d_1\times \vec d_2)|}{\sqrt6}=\frac1{\sqrt6}.6​∣AB⋅(d1​×d2​)∣​=6​1​.

Hence,

∣AB⃗⋅(d⃗1×d⃗2)∣=1.|\vec{AB}\cdot(\vec d_1\times \vec d_2)|=1.∣AB⋅(d1​×d2​)∣=1.

Now,

AB⃗⋅(d⃗1×d⃗2)=(λ−1,2,2)⋅(−1,2,−1).\vec{AB}\cdot(\vec d_1\times \vec d_2) =(\lambda-1,2,2)\cdot(-1,2,-1).AB⋅(d1​×d2​)=(λ−1,2,2)⋅(−1,2,−1).

Compute:

=−(λ−1)+4−2=−λ+1+2=3−λ.=-(\lambda-1)+4-2=-\lambda+1+2=3-\lambda.=−(λ−1)+4−2=−λ+1+2=3−λ.

So,

∣3−λ∣=1.|3-\lambda|=1.∣3−λ∣=1.

Thus,

3−λ=±1.3-\lambda=\pm1.3−λ=±1.

Therefore,

λ1=2,λ2=4.\lambda_1=2,\qquad \lambda_2=4.λ1​=2,λ2​=4.
  1. Form the three points of the circle

The points are

P=(0,0),Q=(λ1,λ2)=(2,4),R=(λ2,λ1)=(4,2).P=(0,0),\qquad Q=(\lambda_1,\lambda_2)=(2,4),\qquad R=(\lambda_2,\lambda_1)=(4,2).P=(0,0),Q=(λ1​,λ2​)=(2,4),R=(λ2​,λ1​)=(4,2).
  1. Find the radius of the circumcircle of triangle PQRPQRPQR

First compute the side lengths:

PQ=(2−0)2+(4−0)2=20=25,PQ=\sqrt{(2-0)^2+(4-0)^2}=\sqrt{20}=2\sqrt5,PQ=(2−0)2+(4−0)2​=20​=25​, PR=(4−0)2+(2−0)2=20=25,PR=\sqrt{(4-0)^2+(2-0)^2}=\sqrt{20}=2\sqrt5,PR=(4−0)2+(2−0)2​=20​=25​, QR=(4−2)2+(2−4)2=4+4=22.QR=\sqrt{(4-2)^2+(2-4)^2}=\sqrt{4+4}=2\sqrt2.QR=(4−2)2+(2−4)2​=4+4​=22​.

Now find area of triangle PQRPQRPQR. Using determinant formula,

Δ=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.\Delta=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.Δ=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣.

For (0,0),(2,4),(4,2)(0,0),(2,4),(4,2)(0,0),(2,4),(4,2),

Δ=12∣0(4−2)+2(2−0)+4(0−4)∣=12∣4−16∣=6.\Delta=\frac12|0(4-2)+2(2-0)+4(0-4)| =\frac12|4-16|=6.Δ=21​∣0(4−2)+2(2−0)+4(0−4)∣=21​∣4−16∣=6.

Circumradius formula:

R=abc4Δ,R=\frac{abc}{4\Delta},R=4Δabc​,

where

a=25,b=25,c=22.a=2\sqrt5,\quad b=2\sqrt5,\quad c=2\sqrt2.a=25​,b=25​,c=22​.

So,

R=(25)(25)(22)4⋅6=8⋅5224=523.R=\frac{(2\sqrt5)(2\sqrt5)(2\sqrt2)}{4\cdot 6} =\frac{8\cdot 5\sqrt2}{24} =\frac{5\sqrt2}{3}.R=4⋅6(25​)(25​)(22​)​=248⋅52​​=352​​.
  1. Compare with the options
R=523R=\frac{5\sqrt2}{3}R=352​​

which matches Option B.

  1. Comparison with stored correct answer

Stored correct answer is B, which agrees with the derived result.

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