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3D Geometry question

2024 · 9 Apr · Shift 1 · Q49
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  5. /2024 · 9 Apr · Shift 1 · Q49

3D Geometry question

2024 · 9 Apr · Shift 1 · Q49

JEE MainMathematics3D GeometryMCQ+4 / −1
Let the line L\mathrm{L}L intersect the lines x−2=−y=z−1,2(x+1)=2(y−1)=z+1x-2=-y=z-1,2(x+1)=2(y-1)=z+1x−2=−y=z−1,2(x+1)=2(y−1)=z+1 and be parallel to the line x−23=y−11=z−22\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}3x−2​=1y−1​=2z−2​. Then which of the following points lies on L\mathrm{L}L ?
  1. A
    (−13,1,−1)\left(-\frac{1}{3}, 1,-1\right)(−31​,1,−1)
  2. B
    (−13,−1,1)\left(-\frac{1}{3},-1,1\right)(−31​,−1,1)
  3. C
    (−13,−1,−1)\left(-\frac{1}{3},-1,-1\right)(−31​,−1,−1)
  4. D
    (−13,1,1)\left(-\frac{1}{3}, 1,1\right)(−31​,1,1)
View written solutionFree

Correct answer: A

  1. Write the given lines in parametric form

The required line LLL:

  • intersects the line x−2=−y=z−1x-2=-y=z-1x−2=−y=z−1
  • intersects the line 2(x+1)=2(y−1)=z+12(x+1)=2(y-1)=z+12(x+1)=2(y−1)=z+1
  • is parallel to the line x−23=y−11=z−22\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}3x−2​=1y−1​=2z−2​

So the direction ratios of LLL are proportional to (3,1,2)(3,1,2)(3,1,2).

  1. First given line

Given: x−2=−y=z−1=tx-2=-y=z-1=tx−2=−y=z−1=t

Hence, x=2+t,y=−t,z=1+tx=2+t,\quad y=-t,\quad z=1+tx=2+t,y=−t,z=1+t

So a general point on the first line is P=(2+t,−t,1+t)P=(2+t,-t,1+t)P=(2+t,−t,1+t)

  1. Second given line

Given: 2(x+1)=2(y−1)=z+1=s2(x+1)=2(y-1)=z+1=s2(x+1)=2(y−1)=z+1=s

Hence, x=−1+s2,y=1+s2,z=−1+sx=-1+\frac{s}{2},\quad y=1+\frac{s}{2},\quad z=-1+sx=−1+2s​,y=1+2s​,z=−1+s

So a general point on the second line is Q=(−1+s2, 1+s2, −1+s)Q=\left(-1+\frac{s}{2},\ 1+\frac{s}{2},\ -1+s\right)Q=(−1+2s​, 1+2s​, −1+s)

  1. Use the fact that LLL is parallel to (3,1,2)(3,1,2)(3,1,2)

Since LLL passes through PPP and QQQ, the vector PQ→\overrightarrow{PQ}PQ​ must be parallel to (3,1,2)(3,1,2)(3,1,2).

Compute: PQ→=Q−P\overrightarrow{PQ}=Q-PPQ​=Q−P

So, PQ→=(−1+s2−(2+t), 1+s2−(−t), −1+s−(1+t))\overrightarrow{PQ}=\left(-1+\frac{s}{2}-(2+t),\ 1+\frac{s}{2}-(-t),\ -1+s-(1+t)\right)PQ​=(−1+2s​−(2+t), 1+2s​−(−t), −1+s−(1+t))

That is, PQ→=(−3+s2−t, 1+s2+t, −2+s−t)\overrightarrow{PQ}=\left(-3+\frac{s}{2}-t,\ 1+\frac{s}{2}+t,\ -2+s-t\right)PQ​=(−3+2s​−t, 1+2s​+t, −2+s−t)

Since this is parallel to (3,1,2)(3,1,2)(3,1,2), there exists kkk such that (−3+s2−t, 1+s2+t, −2+s−t)=k(3,1,2)\left(-3+\frac{s}{2}-t,\ 1+\frac{s}{2}+t,\ -2+s-t\right)=k(3,1,2)(−3+2s​−t, 1+2s​+t, −2+s−t)=k(3,1,2)

Thus, \begin{align*} -3+\frac{s}{2}-t &= 3k \quad ...(1)\ 1+\frac{s}{2}+t &= k \quad ...(2)\ -2+s-t &= 2k \quad ...(3) \end{align*}

  1. Solve for s,t,ks,t,ks,t,k

From (2): k=1+s2+tk=1+\frac{s}{2}+tk=1+2s​+t

Substitute into (3): −2+s−t=2(1+s2+t)-2+s-t=2\left(1+\frac{s}{2}+t\right)−2+s−t=2(1+2s​+t) −2+s−t=2+s+2t-2+s-t=2+s+2t−2+s−t=2+s+2t −4=3t-4=3t−4=3t t=−43t=-\frac{4}{3}t=−34​

Now substitute into (1): −3+s2−(−43)=3k-3+\frac{s}{2}-\left(-\frac{4}{3}\right)=3k−3+2s​−(−34​)=3k −3+s2+43=3k-3+\frac{s}{2}+\frac{4}{3}=3k−3+2s​+34​=3k s2−53=3k\frac{s}{2}-\frac{5}{3}=3k2s​−35​=3k

Also from (2): k=1+s2−43=s2−13k=1+\frac{s}{2}-\frac{4}{3}=\frac{s}{2}-\frac{1}{3}k=1+2s​−34​=2s​−31​

So, s2−53=3(s2−13)\frac{s}{2}-\frac{5}{3}=3\left(\frac{s}{2}-\frac{1}{3}\right)2s​−35​=3(2s​−31​) s2−53=3s2−1\frac{s}{2}-\frac{5}{3}=\frac{3s}{2}-12s​−35​=23s​−1

Multiply by 6: 3s−10=9s−63s-10=9s-63s−10=9s−6 −4=6s-4=6s−4=6s s=−23s=-\frac{2}{3}s=−32​

Then k=s2−13=−13−13=−23k=\frac{s}{2}-\frac{1}{3}=-\frac{1}{3}-\frac{1}{3}=-\frac{2}{3}k=2s​−31​=−31​−31​=−32​

  1. Find the two intersection points

Point on first line: P=(2+t,−t,1+t)P=(2+t,-t,1+t)P=(2+t,−t,1+t) with t=−43t=-\frac{4}{3}t=−34​, P=(2−43,43,1−43)=(23,43,−13)P=\left(2-\frac{4}{3},\frac{4}{3},1-\frac{4}{3}\right)=\left(\frac{2}{3},\frac{4}{3},-\frac{1}{3}\right)P=(2−34​,34​,1−34​)=(32​,34​,−31​)

Point on second line: Q=(−1+s2,1+s2,−1+s)Q=\left(-1+\frac{s}{2},1+\frac{s}{2},-1+s\right)Q=(−1+2s​,1+2s​,−1+s) with s=−23s=-\frac{2}{3}s=−32​, Q=(−1−13,1−13,−1−23)=(−43,23,−53)Q=\left(-1-\frac{1}{3},1-\frac{1}{3},-1-\frac{2}{3}\right)=\left(-\frac{4}{3},\frac{2}{3},-\frac{5}{3}\right)Q=(−1−31​,1−31​,−1−32​)=(−34​,32​,−35​)

  1. Equation of line LLL

Since LLL is parallel to (3,1,2)(3,1,2)(3,1,2) and passes through PPP, its parametric form is x=23+3λ,y=43+λ,z=−13+2λx=\frac{2}{3}+3\lambda,\quad y=\frac{4}{3}+\lambda,\quad z=-\frac{1}{3}+2\lambdax=32​+3λ,y=34​+λ,z=−31​+2λ

Now test the options.

  1. Check option A: (−13,1,−1)\left(-\frac13,1,-1\right)(−31​,1,−1)

From xxx: −13=23+3λ⇒3λ=−1⇒λ=−13-\frac13=\frac23+3\lambda \Rightarrow 3\lambda=-1 \Rightarrow \lambda=-\frac13−31​=32​+3λ⇒3λ=−1⇒λ=−31​

Then y=43−13=1y=\frac43-\frac13=1y=34​−31​=1 z=−13+2(−13)=−13−23=−1z=-\frac13+2\left(-\frac13\right)=-\frac13-\frac23=-1z=−31​+2(−31​)=−31​−32​=−1

So option A lies on LLL.

  1. Check other options quickly

For the same x=−13x=-\frac13x=−31​, we must have λ=−13\lambda=-\frac13λ=−31​. That gives uniquely y=1,z=−1y=1,\quad z=-1y=1,z=−1 So none of B, C, D can lie on LLL.

  1. Final answer

The point lying on LLL is (−13,1,−1)\boxed{\left(-\frac13,1,-1\right)}(−31​,1,−1)​ which is Option A.

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