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

2025 · 22 Jan · Shift 1 · Q48
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. /2025 · 22 Jan · Shift 1 · Q48

3D Geometry question

2025 · 22 Jan · Shift 1 · Q48

JEE MainMathematics3D GeometryNumerical+4 / −1
Let L1:x−13=y−1−1=z+10\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}L1​:3x−1​=−1y−1​=0z+1​ and L2:x−22=y0=z+4α,α∈R\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}L2​:2x−2​=0y​=αz+4​,α∈R, be two lines, which intersect at the point BBB. If PPP is the foot of perpendicular from the point A(1,1,−1)A(1,1,-1)A(1,1,−1) on L2L_2L2​, then the value of 26α( PB)226 \alpha(\mathrm{~PB})^226α( PB)2 is ‾\underline{\hspace{2cm}}​ .
Numerical answer
View written solutionFree

Correct answer: 216

  1. Write the parametric equations of the two lines

For L1L_1L1​:

x−13=y−1−1=z+10=t\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=t3x−1​=−1y−1​=0z+1​=t

So,

x=1+3t,y=1−t,z=−1x=1+3t,\quad y=1-t,\quad z=-1x=1+3t,y=1−t,z=−1

Thus L1L_1L1​ lies in the plane z=−1z=-1z=−1.

For L2L_2L2​:

x−22=y0=z+4α=s\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}=s2x−2​=0y​=αz+4​=s

So,

x=2+2s,y=0,z=−4+αsx=2+2s,\quad y=0,\quad z=-4+\alpha sx=2+2s,y=0,z=−4+αs
  1. Use the condition that the lines intersect at point BBB

At the intersection, coordinates must match for some t,st,st,s.

From the yyy-coordinate:

1−t=0  ⟹  t=11-t=0 \implies t=11−t=0⟹t=1

Then from the xxx-coordinate:

1+3(1)=2+2s1+3(1)=2+2s1+3(1)=2+2s 4=2+2s  ⟹  s=14=2+2s \implies s=14=2+2s⟹s=1

Now from the zzz-coordinate:

−1=−4+α(1)-1=-4+\alpha(1)−1=−4+α(1) α=3\alpha=3α=3

So,

α=3\alpha=3α=3

Also, the intersection point is

B=(4,0,−1)B=(4,0,-1)B=(4,0,−1)
  1. Find the foot of perpendicular PPP from A(1,1,−1)A(1,1,-1)A(1,1,−1) to L2L_2L2​

Now L2L_2L2​ becomes

x=2+2s,y=0,z=−4+3sx=2+2s,\quad y=0,\quad z=-4+3sx=2+2s,y=0,z=−4+3s

A point on L2L_2L2​ is

Q(2,0,−4)Q(2,0,-4)Q(2,0,−4)

with direction vector

d⃗=(2,0,3)\vec d=(2,0,3)d=(2,0,3)

Let the foot PPP correspond to parameter s=λs=\lambdas=λ. Then

P=(2+2λ,0,−4+3λ)P=(2+2\lambda,0,-4+3\lambda)P=(2+2λ,0,−4+3λ)

Since AP⊥L2AP \perp L_2AP⊥L2​, we use

(AP⃗)⋅d⃗=0(\vec{AP})\cdot \vec d=0(AP)⋅d=0

Now,

AP⃗=P−A=(2+2λ−1,0−1,−4+3λ+1)\vec{AP}=P-A=(2+2\lambda-1,0-1,-4+3\lambda+1)AP=P−A=(2+2λ−1,0−1,−4+3λ+1) AP⃗=(1+2λ,−1,−3+3λ)\vec{AP}=(1+2\lambda,-1,-3+3\lambda)AP=(1+2λ,−1,−3+3λ)

Dot product with (2,0,3)(2,0,3)(2,0,3):

(1+2λ)2+(−1)0+(−3+3λ)3=0(1+2\lambda)2+(-1)0+(-3+3\lambda)3=0(1+2λ)2+(−1)0+(−3+3λ)3=0 2+4λ−9+9λ=02+4\lambda-9+9\lambda=02+4λ−9+9λ=0 13λ−7=013\lambda-7=013λ−7=0 λ=713\lambda=\frac{7}{13}λ=137​

Hence,

P=(2+1413,0,−4+2113)P=\left(2+\frac{14}{13},0,-4+\frac{21}{13}\right)P=(2+1314​,0,−4+1321​) P=(4013,0,−3113)P=\left(\frac{40}{13},0,-\frac{31}{13}\right)P=(1340​,0,−1331​)
  1. Compute PBPBPB

Point BBB corresponds to s=1s=1s=1 on L2L_2L2​. So along the line,

PB=∣1−λ∣ ∣d⃗∣PB=|1-\lambda|\,|\vec d|PB=∣1−λ∣∣d∣

with

∣d⃗∣=22+02+32=13|\vec d|=\sqrt{2^2+0^2+3^2}=\sqrt{13}∣d∣=22+02+32​=13​

Thus,

PB=∣1−713∣13=61313PB=\left|1-\frac{7}{13}\right|\sqrt{13}=\frac{6}{13}\sqrt{13}PB=​1−137​​13​=136​13​

Therefore,

(PB)2=36169⋅13=3613(PB)^2=\frac{36}{169}\cdot 13=\frac{36}{13}(PB)2=16936​⋅13=1336​
  1. Find 26α(PB)226\alpha (PB)^226α(PB)2

Since α=3\alpha=3α=3,

26α(PB)2=26⋅3⋅361326\alpha (PB)^2=26\cdot 3\cdot \frac{36}{13}26α(PB)2=26⋅3⋅1336​ =2⋅3⋅36=2\cdot 3\cdot 36=2⋅3⋅36 =216=216=216
  1. Compare with stored answer

Derived answer: 216216216

Stored correct answer: 216216216

They match.

PreviousNext

More from 3D Geometry

  • Let a line pass through two distinct points P(−2,−1,3) and Q, and be parallel to the vector 3i^+2j^​+2k^. If the distance of the point Q from the point R(1,3,3) is 5 , then the square of the area of △PQR…2025 · MCQ
  • The perpendicular distance, of the line 2x−1​=−1y+2​=2z+3​ from the point P(2,−10,1), is :2025 · MCQ
  • Let P be the foot of the perpendicular from the point Q(10,−3,−1) on the line 7x−3​=−1y−2​=−2z+1​. Then the area of the right angled triangle PQR, where R is the point (3,−2,1), is2025 · MCQ
  • If the square of the shortest distance between the lines 1x−2​=2y−1​=−3z+3​ and 2x+1​=4y+3​=−5z+5​ is nm​, where m, n are coprime numbers, then m+n is equal to :2025 · MCQ
  • The distance of the line 2x−2​=3y−6​=4z−3​ from the point (1,4,0) along the line 1x​=2y−2​=3z+3​ is :2025 · MCQ
  • Let in a △ABC, the length of the side AC be 6 , the vertex B be (1,2,3) and the vertices A,C lie on the line 3x−6​=2y−7​=−2z−7​. Then the area (in sq. units) of △ABC is:2025 · MCQ
  • Let the line passing through the points (−1,2,1) and parallel to the line 2x−1​=3y+1​=4z​ intersect the line 3x+2​=2y−3​=1z−4​ at the point P. Then the distance of P from the point Q(4,−5,1)…2025 · MCQ
  • Let P be the image of the point Q(7,−2,5) in the line L:2x−1​=3y+1​=4z​ and R(5,p,q) be a point on L. Then the square of the area of △PQR is ​…2025 · Numerical