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3D Geometry question

2021 · 24 Feb · Shift 2 · Q36
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  5. /2021 · 24 Feb · Shift 2 · Q36

3D Geometry question

2021 · 24 Feb · Shift 2 · Q36

JEE MainMathematics3D GeometryNumerical+4 / −1
Let λ\lambdaλ be an integer. If the shortest distance between the lines x −λ-\lambda−λ= 2y −-− 1 =−-− 2z and x = y + 2 λ\lambdaλ= z −λ-\lambda−λ is 722{{\sqrt 7 } \over {2\sqrt 2 }}22​7​​, then the value of | λ\lambdaλ | is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1

Let the two lines be written in symmetric/parametric form.

1. Write the lines in parametric form

Line L1L_1L1​

Given x−λ=2y−1=−2zx-\lambda = 2y-1 = -2zx−λ=2y−1=−2z Let the common value be ttt. Then x−λ=t,2y−1=t,−2z=tx-\lambda=t,\quad 2y-1=t,\quad -2z=tx−λ=t,2y−1=t,−2z=t So, x=λ+t,y=t+12,z=−t2x=\lambda+t,\qquad y=\frac{t+1}{2},\qquad z=-\frac t2x=λ+t,y=2t+1​,z=−2t​ Hence a point on L1L_1L1​ is A(λ,12,0)A\left(\lambda,\frac12,0\right)A(λ,21​,0) and a direction vector is d⃗1=(1,12,−12)\vec d_1=(1,\tfrac12,-\tfrac12)d1​=(1,21​,−21​) We may multiply by 222 to simplify: d⃗1=(2,1,−1)\vec d_1=(2,1,-1)d1​=(2,1,−1)

Line L2L_2L2​

Given x=y+2λ=z−λx = y+2\lambda = z-\lambdax=y+2λ=z−λ Let the common value be sss. Then x=s,y=s−2λ,z=s+λx=s,\quad y=s-2\lambda,\quad z=s+\lambdax=s,y=s−2λ,z=s+λ Hence a point on L2L_2L2​ is B(0,−2λ,λ)B(0,-2\lambda,\lambda)B(0,−2λ,λ) and a direction vector is d⃗2=(1,1,1)\vec d_2=(1,1,1)d2​=(1,1,1)

2. 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​, 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​)∣​ where AB→=B−A\overrightarrow{AB}=B-AAB=B−A

Here, A(λ,12,0),B(0,−2λ,λ)A\left(\lambda,\frac12,0\right),\qquad B(0,-2\lambda,\lambda)A(λ,21​,0),B(0,−2λ,λ) So AB→=(−λ,−2λ−12,λ)\overrightarrow{AB}=(-\lambda,-2\lambda-\tfrac12,\lambda)AB=(−λ,−2λ−21​,λ)

3. Compute d⃗1×d⃗2\vec d_1\times \vec d_2d1​×d2​

Take d⃗1=(2,1,−1),d⃗2=(1,1,1)\vec d_1=(2,1,-1),\qquad \vec d_2=(1,1,1)d1​=(2,1,−1),d2​=(1,1,1) Then

\begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & -1\\ 1 & 1 & 1 \end{vmatrix}$$ $$=\hat i(1\cdot1-(-1)\cdot1)-\hat j(2\cdot1-(-1)\cdot1)+\hat k(2\cdot1-1\cdot1)$$ $$=(2,-3,1)$$ Thus, $$|\vec d_1\times \vec d_2|=\sqrt{2^2+(-3)^2+1^2}=\sqrt{14}$$ --- ## 4. Compute the scalar triple product $$\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2) =(-\lambda,-2\lambda-\tfrac12,\lambda)\cdot(2,-3,1)$$ $$=-2\lambda+(-2\lambda-\tfrac12)(-3)+\lambda$$ $$=-2\lambda+6\lambda+\tfrac32+\lambda$$ $$=5\lambda+\tfrac32$$ Therefore, $$D=\frac{|5\lambda+\tfrac32|}{\sqrt{14}}$$ Given shortest distance is $$\frac{\sqrt7}{2\sqrt2}$$ Note that $$\frac{\sqrt7}{2\sqrt2}=\frac{\sqrt{14}}{4}$$ So, $$\frac{|5\lambda+\tfrac32|}{\sqrt{14}}=\frac{\sqrt{14}}{4}$$ Multiply by $\sqrt{14}$: $$|5\lambda+\tfrac32|=\frac{14}{4}=\frac72$$ Thus, $$\left|10\lambda+3\right|=7$$ So, $$10\lambda+3=\pm 7$$ ### Case 1: $$10\lambda+3=7$$ $$10\lambda=4$$ $$\lambda=\frac25$$ Not an integer. ### Case 2: $$10\lambda+3=-7$$ $$10\lambda=-10$$ $$\lambda=-1$$ This is an integer. Hence, $$|\lambda|=1$$ ## 5. Comparison with stored answer Stored correct answer: $1$ Our derived answer is also $1$, so it agrees.
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