Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

3D Geometry question

2024 · 30 Jan · Shift 2 · Q55
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /3D Geometry
  5. /2024 · 30 Jan · Shift 2 · Q55

3D Geometry question

2024 · 30 Jan · Shift 2 · Q55

JEE MainMathematics3D GeometryNumerical+4 / −1
Let a line passing through the point (−1,2,3)(-1,2,3)(−1,2,3) intersect the lines L1:x−13=y−22=z+1−2L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}L1​:3x−1​=2y−2​=−2z+1​ at M(α,β,γ)M(\alpha, \beta, \gamma)M(α,β,γ) and L2:x+2−3=y−2−2=z−14L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}L2​:−3x+2​=−2y−2​=4z−1​ at N(a,b,c)N(a, b, c)N(a,b,c). Then, the value of (α+β+γ)2(a+b+c)2\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}(a+b+c)2(α+β+γ)2​ equals ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 196

  1. Write the two given lines in parametric form

For L1L_1L1​:

x−13=y−22=z+1−2=t\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}=t3x−1​=2y−2​=−2z+1​=t

So a general point on L1L_1L1​ is

M(1+3t,  2+2t,  −1−2t).M(1+3t,\; 2+2t,\; -1-2t).M(1+3t,2+2t,−1−2t).

For L2L_2L2​:

x+2−3=y−2−2=z−14=s\frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}=s−3x+2​=−2y−2​=4z−1​=s

So a general point on L2L_2L2​ is

N(−2−3s,  2−2s,  1+4s).N(-2-3s,\; 2-2s,\; 1+4s).N(−2−3s,2−2s,1+4s).

The given point through which the required line passes is

P(−1,2,3).P(-1,2,3).P(−1,2,3).
  1. Use collinearity of M,P,NM, P, NM,P,N

Since the required line passes through PPP and intersects L1L_1L1​ at MMM and L2L_2L2​ at NNN, the points M,P,NM, P, NM,P,N are collinear.

So vectors PM→\overrightarrow{PM}PM and PN→\overrightarrow{PN}PN must be parallel.

Compute:

PM→=M−P=(1+3t+1,  2+2t−2,  −1−2t−3)\overrightarrow{PM}=M-P=(1+3t+1,\; 2+2t-2,\; -1-2t-3)PM=M−P=(1+3t+1,2+2t−2,−1−2t−3) PM→=(2+3t,  2t,  −4−2t)\overrightarrow{PM}=(2+3t,\; 2t,\; -4-2t)PM=(2+3t,2t,−4−2t)

Similarly,

PN→=N−P=(−2−3s+1,  2−2s−2,  1+4s−3)\overrightarrow{PN}=N-P=(-2-3s+1,\; 2-2s-2,\; 1+4s-3)PN=N−P=(−2−3s+1,2−2s−2,1+4s−3) PN→=(−1−3s,  −2s,  −2+4s)\overrightarrow{PN}=(-1-3s,\; -2s,\; -2+4s)PN=(−1−3s,−2s,−2+4s)

Thus,

(2+3t,  2t,  −4−2t)=λ(−1−3s,  −2s,  −2+4s)(2+3t,\; 2t,\; -4-2t)=\lambda(-1-3s,\; -2s,\; -2+4s)(2+3t,2t,−4−2t)=λ(−1−3s,−2s,−2+4s)

for some scalar λ\lambdaλ.

This gives:

2+3t=λ(−1−3s)(1)2+3t=\lambda(-1-3s) \quad (1)2+3t=λ(−1−3s)(1) 2t=λ(−2s)(2)2t=\lambda(-2s) \quad (2)2t=λ(−2s)(2) −4−2t=λ(−2+4s)(3)-4-2t=\lambda(-2+4s) \quad (3)−4−2t=λ(−2+4s)(3)
  1. Eliminate λ\lambdaλ

From (2):

λ=−ts(s≠0)\lambda=-\frac{t}{s} \quad (s\neq 0)λ=−st​(s=0)

Substitute into (1):

2+3t=(−ts)(−1−3s)=ts(1+3s)2+3t=\left(-\frac{t}{s}\right)(-1-3s)=\frac{t}{s}(1+3s)2+3t=(−st​)(−1−3s)=st​(1+3s)

Multiply by sss:

s(2+3t)=t(1+3s)s(2+3t)=t(1+3s)s(2+3t)=t(1+3s) 2s+3st=t+3st2s+3st=t+3st2s+3st=t+3st 2s=t2s=t2s=t

So,

t=2s.(4)t=2s. \quad (4)t=2s.(4)

Now use (3):

−4−2t=(−ts)(−2+4s)=ts(2−4s)-4-2t=\left(-\frac{t}{s}\right)(-2+4s)=\frac{t}{s}(2-4s)−4−2t=(−st​)(−2+4s)=st​(2−4s)

Substitute t=2st=2st=2s:

−4−4s=2ss(2−4s)=2(2−4s)=4−8s-4-4s=\frac{2s}{s}(2-4s)=2(2-4s)=4-8s−4−4s=s2s​(2−4s)=2(2−4s)=4−8s −4−4s=4−8s-4-4s=4-8s−4−4s=4−8s 4s=84s=84s=8 s=2s=2s=2

Hence,

t=2s=4.t=2s=4.t=2s=4.
  1. Find the intersection points MMM and NNN

For L1L_1L1​, with t=4t=4t=4:

M=(1+3⋅4,  2+2⋅4,  −1−2⋅4)=(13,10,−9)M=(1+3\cdot 4,\; 2+2\cdot 4,\; -1-2\cdot 4)=(13,10,-9)M=(1+3⋅4,2+2⋅4,−1−2⋅4)=(13,10,−9)

So,

α=13,  β=10,  γ=−9.\alpha=13,\; \beta=10,\; \gamma=-9.α=13,β=10,γ=−9.

Then

α+β+γ=13+10−9=14.\alpha+\beta+\gamma=13+10-9=14.α+β+γ=13+10−9=14.

For L2L_2L2​, with s=2s=2s=2:

N=(−2−3⋅2,  2−2⋅2,  1+4⋅2)=(−8,−2,9)N=(-2-3\cdot 2,\; 2-2\cdot 2,\; 1+4\cdot 2)=(-8,-2,9)N=(−2−3⋅2,2−2⋅2,1+4⋅2)=(−8,−2,9)

So,

a=−8,  b=−2,  c=9.a=-8,\; b=-2,\; c=9.a=−8,b=−2,c=9.

Then

a+b+c=−8−2+9=−1.a+b+c=-8-2+9=-1.a+b+c=−8−2+9=−1.
  1. Compute the required value
(α+β+γ)2(a+b+c)2=142(−1)2=196.\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2} =\frac{14^2}{(-1)^2}=196.(a+b+c)2(α+β+γ)2​=(−1)2142​=196.
  1. Comparison with stored answer

Derived answer = 196196196.

Stored correct answer = 196196196.

So the answer matches.

PreviousNext

More from 3D Geometry

  • Let Q and R be the feet of perpendiculars from the point P(a,a,a) on the lines x=y,z=1 and x=−y,z=−1 respectively. If ∠QPR is a right angle, then 12a2 is equal to ​…2024 · Numerical
  • Let (α,β,γ) be the mirror image of the point (2,3,5) in the line 2x−1​=3y−2​=4z−3​. Then, 2α+3β+4γ is equal to2024 · MCQ
  • The shortest distance, between lines L1​ and L2​, where L1​:2x−1​=−3y+1​=2z+4​ and L2​ is the line, passing through the points A(−4,4,3),B(−1,6,3) and perpendicular to the line −2x−3​=3y​=1z−1​…2024 · MCQ
  • A line passes through A(4,−6,−2) and B(16,−2,4). The point P(a,b,c), where a,b,c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points P(a,b,c) and Q(4,−12,3)…2024 · Numerical
  • The shortest distance between the lines 1x−5​=2y−2​=−3z−4​ and 1x+3​=4y+5​=−5z−1​ is :2023 · MCQ
  • One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges along x,y and z axes are 3,4 and 5 units respectively. Let P be the vertex (3,4,5). Then the shortest distance…2023 · MCQ
  • If the lines 2x−1​=−32−y​=αz−3​ and 5x−4​=2y−1​=βz​ intersect, then the magnitude of the minimum value of 8αβ is ​.2023 · Numerical
  • The shortest distance between the lines 4x−4​=5y+2​=3z+3​ and 3x−1​=4y−3​=2z−4​ is :2023 · MCQ