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3D Geometry question

2021 · 22 Jul · Shift 2 · Q30
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  5. /2021 · 22 Jul · Shift 2 · Q30

3D Geometry question

2021 · 22 Jul · Shift 2 · Q30

JEE MainMathematics3D GeometryMCQ+4 / −1
If the shortest distance between the straight lines 3(x−1)=6(y−2)=2(z−1)3(x - 1) = 6(y - 2) = 2(z - 1)3(x−1)=6(y−2)=2(z−1) and 4(x−2)=2(y−λ)=(z−3),λ∈R4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R4(x−2)=2(y−λ)=(z−3),λ∈R is 138{1 \over {\sqrt {38} }}38​1​, then the integral value of λ\lambdaλ is equal to :
  1. A
    3
  2. B
    2
  3. C
    5
  4. D
    −-− 1
View written solutionFree

Correct answer: A

  1. Write the lines in symmetric/parametric form

For the first line, 3(x−1)=6(y−2)=2(z−1)=t3(x-1)=6(y-2)=2(z-1)=t3(x−1)=6(y−2)=2(z−1)=t So x=1+t3,y=2+t6,z=1+t2x=1+\frac t3,\quad y=2+\frac t6,\quad z=1+\frac t2x=1+3t​,y=2+6t​,z=1+2t​ Hence a point on the line is A=(1,2,1)A=(1,2,1)A=(1,2,1) and its direction vector is d⃗1=(13,16,12)∝(2,1,3).\vec d_1=\left(\frac13,\frac16,\frac12\right)\propto (2,1,3).d1​=(31​,61​,21​)∝(2,1,3). So we take d⃗1=(2,1,3).\vec d_1=(2,1,3).d1​=(2,1,3).

For the second line, 4(x−2)=2(y−λ)=z−3=s4(x-2)=2(y-\lambda)=z-3=s4(x−2)=2(y−λ)=z−3=s So x=2+s4,y=λ+s2,z=3+sx=2+\frac s4,\quad y=\lambda+\frac s2,\quad z=3+sx=2+4s​,y=λ+2s​,z=3+s Hence a point on the line is B=(2,λ,3)B=(2,\lambda,3)B=(2,λ,3) and its direction vector is d⃗2=(14,12,1)∝(1,2,4).\vec d_2=\left(\frac14,\frac12,1\right)\propto (1,2,4).d2​=(41​,21​,1)∝(1,2,4). So we take d⃗2=(1,2,4).\vec d_2=(1,2,4).d2​=(1,2,4).

  1. Formula for shortest distance between two skew lines

If the lines are r⃗=a⃗+td⃗1,r⃗=b⃗+sd⃗2,\vec r=\vec a+t\vec d_1,\qquad \vec r=\vec b+s\vec d_2,r=a+td1​,r=b+sd2​, then the shortest distance 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=(2−1,λ−2,3−1)=(1,λ−2,2).\vec{AB}=B-A=(2-1,\lambda-2,3-1)=(1,\lambda-2,2).AB=B−A=(2−1,λ−2,3−1)=(1,λ−2,2).

  1. Compute the cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & 3\\ 1 & 2 & 4 \end{vmatrix}$$ $$=\hat i(1\cdot 4-3\cdot 2)-\hat j(2\cdot 4-3\cdot 1)+\hat k(2\cdot 2-1\cdot 1)$$ $$=(-2,-5,3).$$ So, $$|\vec d_1\times \vec d_2|=\sqrt{(-2)^2+(-5)^2+3^2}=\sqrt{38}.$$ --- 4. **Use the given shortest distance** Given $$D=\frac1{\sqrt{38}}.$$ Thus, $$\frac{|\vec{AB}\cdot (\vec d_1\times \vec d_2)|}{\sqrt{38}}=\frac1{\sqrt{38}}$$ which gives $$|\vec{AB}\cdot (\vec d_1\times \vec d_2)|=1.$$ Now, $$\vec{AB}\cdot (\vec d_1\times \vec d_2)=(1,\lambda-2,2)\cdot (-2,-5,3)$$ $$=-2-5(\lambda-2)+6$$ $$=-2-5\lambda+10+6$$ $$=14-5\lambda.$$ Therefore, $$|14-5\lambda|=1.$$ So, $$14-5\lambda=\pm 1.$$ Case 1: $$14-5\lambda=1\implies 5\lambda=13\implies \lambda=\frac{13}{5}.$$ Case 2: $$14-5\lambda=-1\implies 5\lambda=15\implies \lambda=3.$$ The **integral** value is $$\boxed{3}.$$ --- 5. **Check with options** - A: $3$ ✅ - B: $2$ ❌ - C: $5$ ❌ - D: $-1$ ❌ So the correct option is **A**.
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