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3D Geometry question

2024 · 31 Jan · Shift 2 · Q38
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  5. /2024 · 31 Jan · Shift 2 · Q38

3D Geometry question

2024 · 31 Jan · Shift 2 · Q38

JEE MainMathematics3D GeometryMCQ+4 / −1
The shortest distance, between lines L1L_1L1​ and L2L_2L2​, where L1:x−12=y+1−3=z+42L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}L1​:2x−1​=−3y+1​=2z+4​ and L2L_2L2​ is the line, passing through the points A(−4,4,3),B(−1,6,3)\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)A(−4,4,3),B(−1,6,3) and perpendicular to the line x−3−2=y3=z−11\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}−2x−3​=3y​=1z−1​, is
  1. A
    141221\frac{141}{\sqrt{221}}221​141​
  2. B
    24117\frac{24}{\sqrt{117}}117​24​
  3. C
    42117\frac{42}{\sqrt{117}}117​42​
  4. D
    121221\frac{121}{\sqrt{221}}221​121​
View written solutionFree

Correct answer: A

  1. Write line L1L_1L1​ in vector form

Given

L1:x−12=y+1−3=z+42L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}L1​:2x−1​=−3y+1​=2z+4​

So a point on L1L_1L1​ is

P1=(1,−1,−4)P_1=(1,-1,-4)P1​=(1,−1,−4)

and its direction vector is

d⃗1=(2,−3,2).\vec d_1=(2,-3,2).d1​=(2,−3,2).
  1. Find the equation of line L2L_2L2​

We are told L2L_2L2​ passes through points A(−4,4,3)A(-4,4,3)A(−4,4,3) and B(−1,6,3)B(-1,6,3)B(−1,6,3). So its direction vector is

AB→=(−1+4,6−4,3−3)=(3,2,0).\overrightarrow{AB}=(-1+4,6-4,3-3)=(3,2,0).AB=(−1+4,6−4,3−3)=(3,2,0).

Hence L2L_2L2​ is the line through AAA with direction

d⃗2=(3,2,0).\vec d_2=(3,2,0).d2​=(3,2,0).

Let us verify the perpendicularity condition with the line

x−3−2=y3=z−11.\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}.−2x−3​=3y​=1z−1​.

Its direction vector is

(−2,3,1).(-2,3,1).(−2,3,1).

Now,

(3,2,0)⋅(−2,3,1)=−6+6+0=0,(3,2,0)\cdot(-2,3,1)=-6+6+0=0,(3,2,0)⋅(−2,3,1)=−6+6+0=0,

so indeed ABABAB is perpendicular to the given line. Thus L2L_2L2​ is exactly the line through A,BA,BA,B.

  1. Formula for shortest distance between two skew lines

For lines with points P1,P2P_1,P_2P1​,P2​ and direction vectors d⃗1,d⃗2\vec d_1,\vec d_2d1​,d2​, the shortest distance is

D=∣(P1P2→)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.D=\frac{|(\overrightarrow{P_1P_2})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.D=∣d1​×d2​∣∣(P1​P2​​)⋅(d1​×d2​)∣​.

Take a point on L2L_2L2​ as

P2=A=(−4,4,3).P_2=A=(-4,4,3).P2​=A=(−4,4,3).

Then

P1P2→=(−4−1, 4−(−1), 3−(−4))=(−5,5,7).\overrightarrow{P_1P_2}=(-4-1,\ 4-(-1),\ 3-(-4))=(-5,5,7).P1​P2​​=(−4−1, 4−(−1), 3−(−4))=(−5,5,7).
  1. Compute d⃗1×d⃗2\vec d_1\times \vec d_2d1​×d2​
d⃗1×d⃗2=∣i^j^k^2−32320∣\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & -3 & 2\\ 3 & 2 & 0 \end{vmatrix}d1​×d2​=​i^23​j^​−32​k^20​​ =i^((−3)(0)−2⋅2)−j^(2⋅0−2⋅3)+k^(2⋅2−(−3)⋅3)=\hat i((-3)(0)-2\cdot 2)-\hat j(2\cdot 0-2\cdot 3)+\hat k(2\cdot 2-(-3)\cdot 3)=i^((−3)(0)−2⋅2)−j^​(2⋅0−2⋅3)+k^(2⋅2−(−3)⋅3) =(−4,6,13).=(-4,6,13).=(−4,6,13).

Its magnitude is

∣d⃗1×d⃗2∣=(−4)2+62+132=16+36+169=221.|\vec d_1\times \vec d_2|=\sqrt{(-4)^2+6^2+13^2}= \sqrt{16+36+169}=\sqrt{221}.∣d1​×d2​∣=(−4)2+62+132​=16+36+169​=221​.
  1. Compute the scalar triple product
P1P2→⋅(d⃗1×d⃗2)=(−5,5,7)⋅(−4,6,13)\overrightarrow{P_1P_2}\cdot(\vec d_1\times \vec d_2) =(-5,5,7)\cdot(-4,6,13)P1​P2​​⋅(d1​×d2​)=(−5,5,7)⋅(−4,6,13) =20+30+91=141.=20+30+91=141.=20+30+91=141.

Therefore,

D=141221.D=\frac{141}{\sqrt{221}}.D=221​141​.
  1. Match with options
141221\boxed{\frac{141}{\sqrt{221}}}221​141​​

which is Option A.

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So the stored answer is correct.

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