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3D Geometry question

2022 · 24 Jun · Shift 2 · Q33
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  5. /2022 · 24 Jun · Shift 2 · Q33

3D Geometry question

2022 · 24 Jun · Shift 2 · Q33

JEE MainMathematics3D GeometryMCQ+4 / −1
If the shortest distance between the lines x−12=y−23=z−3λ{{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }2x−1​=3y−2​=λz−3​ and x−21=y−44=z−55{{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}1x−2​=4y−4​=5z−5​ is 13{1 \over {\sqrt 3 }}3​1​, then the sum of all possible value of λ\lambdaλ is :
  1. A
    16
  2. B
    6
  3. C
    12
  4. D
    15
View written solutionFree

Correct answer: A

  1. Write the two lines in vector form

The given lines are

x−12=y−23=z−3λ\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}2x−1​=3y−2​=λz−3​ and x−21=y−44=z−55.\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}.1x−2​=4y−4​=5z−5​.

So we can identify:

  • A point on the first line: A(1,2,3)A(1,2,3)A(1,2,3)

  • Direction vector of the first line: d⃗1=(2,3,λ)\vec d_1=(2,3,\lambda)d1​=(2,3,λ)

  • A point on the second line: B(2,4,5)B(2,4,5)B(2,4,5)

  • Direction vector of the second line: d⃗2=(1,4,5)\vec d_2=(1,4,5)d2​=(1,4,5)

  1. Formula for shortest distance between two skew lines

The shortest distance between two lines is

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

where

AB⃗=AB→=B−A=(2−1,4−2,5−3)=(1,2,2).\vec{AB}=\overrightarrow{AB}=B-A=(2-1,4-2,5-3)=(1,2,2).AB=AB=B−A=(2−1,4−2,5−3)=(1,2,2).

Given that

D=13.D=\frac{1}{\sqrt3}.D=3​1​.

  1. Compute the cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 3 & \lambda\\ 1 & 4 & 5 \end{vmatrix}$$ $$=\hat i(15-4\lambda)-\hat j(10-\lambda)+\hat k(8-3)$$ $$=(15-4\lambda,\,\lambda-10,\,5).$$ 4. **Compute the scalar triple product** $$\vec{AB}\cdot(\vec d_1\times \vec d_2) =(1,2,2)\cdot(15-4\lambda,\lambda-10,5)$$ $$=(15-4\lambda)+2(\lambda-10)+2(5)$$ $$=15-4\lambda+2\lambda-20+10$$ $$=5-2\lambda.$$ So, $$|(\vec{AB})\cdot(\vec d_1\times \vec d_2)|=|5-2\lambda|.$$ 5. **Compute the magnitude of the cross product** $$|\vec d_1\times \vec d_2|=\sqrt{(15-4\lambda)^2+(\lambda-10)^2+5^2}.$$ Hence, $$\frac{|5-2\lambda|}{\sqrt{(15-4\lambda)^2+(\lambda-10)^2+25}}=\frac{1}{\sqrt3}.$$ 6. **Solve the equation** Squaring both sides, $$\frac{(5-2\lambda)^2}{(15-4\lambda)^2+(\lambda-10)^2+25}=\frac13.$$ Therefore, $$3(5-2\lambda)^2=(15-4\lambda)^2+(\lambda-10)^2+25.$$ Now expand both sides. Left side: $$3(25-20\lambda+4\lambda^2)=75-60\lambda+12\lambda^2.$$ Right side: $$(15-4\lambda)^2=225-120\lambda+16\lambda^2$$ $$(\lambda-10)^2=\lambda^2-20\lambda+100$$ So, $$225-120\lambda+16\lambda^2+\lambda^2-20\lambda+100+25$$ $$=17\lambda^2-140\lambda+350.$$ Thus, $$75-60\lambda+12\lambda^2=17\lambda^2-140\lambda+350.$$ Bring all terms to one side: $$0=5\lambda^2-80\lambda+275$$ $$\lambda^2-16\lambda+55=0.$$ Factorizing, $$(\lambda-11)(\lambda-5)=0.$$ So, $$\lambda=11 \quad \text{or} \quad \lambda=5.$$ 7. **Find the sum of all possible values** $$11+5=16.$$ 8. **Check with options** Option A is $16$. So the correct answer is **A**.
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