Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let the lines L_ 1 : r = i +2 j +3 k +λ(2 i +3 j +4 k ), λ ∈ ℝ and...

Let the lines L1:r⃗=i^+2j^+3k^+λ(2i^+3j^+4k^),λ∈R\mathrm{L}_{1}: \vec{r}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}), \lambda \in \mathbb{R} and L2:r⃗=(4i^+j^)+μ(5i^+2j^+k^),μ∈R\mathrm{L}_{2}: \vec{r}=(4 \hat{\mathrm{i}}+\hat{\mathrm{j}})+\mu(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \mu \in \mathbb{R}, intersect at the point R. Let P and Q be the points lying on lines L1\mathrm{L}_{1} and L2\mathrm{L}_{2}, respectively, such that ∣PR→∣=29|\overrightarrow{\mathrm{PR}}|=\sqrt{29} and ∣PQ→∣=473|\overrightarrow{\mathrm{PQ}}|=\sqrt{\frac{47}{3}}. If the point P lies in the first octant, then 27(QR)227(\mathrm{QR})^{2} is equal to

Options

  1. A348
  2. B340
  3. C320
  4. D360

Correct answer

D. 360

Step-by-step solution

Find R by solving L1=L2L_1 = L_2:
From 1+2λ=4+5μ1+2\lambda = 4+5\mu, 2+3λ=1+2μ2+3\lambda = 1+2\mu, 3+4λ=μ3+4\lambda = \mu we get λ=−1\lambda = -1, μ=−1\mu = -1, so R=(−1,−1,−1)R = (-1, -1, -1)
Point P on L1L_1: P=(1+2λP,2+3λP,3+4λP)P = (1+2\lambda_P, 2+3\lambda_P, 3+4\lambda_P) with ∣PR→∣2=29|\overrightarrow{PR}|^2 = 29
4(1+λP)2+9(1+λP)2+16(1+λP)2=294(1+\lambda_P)^2 + 9(1+\lambda_P)^2 + 16(1+\lambda_P)^2 = 29 → 29(1+λP)2=2929(1+\lambda_P)^2 = 29 → λP=0\lambda_P = 0 (first octant)
Thus P=(1,2,3)P = (1, 2, 3)
Point Q on L2L_2: Q=(4+5μQ,1+2μQ,μQ)Q = (4+5\mu_Q, 1+2\mu_Q, \mu_Q) with ∣PQ→∣2=473|\overrightarrow{PQ}|^2 = \frac{47}{3}
(3+5μQ)2+(−1+2μQ)2+(μQ−3)2=473(3+5\mu_Q)^2 + (-1+2\mu_Q)^2 + (\mu_Q-3)^2 = \frac{47}{3}
30μQ2+20μQ+19=47330\mu_Q^2 + 20\mu_Q + 19 = \frac{47}{3} → 9μQ2+6μQ+1=09\mu_Q^2 + 6\mu_Q + 1 = 0 → μQ=−13\mu_Q = -\frac{1}{3}
So Q=(73,13,−13)Q = (\frac{7}{3}, \frac{1}{3}, -\frac{1}{3})
∣QR∣2=100+16+49=1209=403|QR|^2 = \frac{100+16+4}{9} = \frac{120}{9} = \frac{40}{3}
27(QR)2=36027(QR)^2 = 360
Practice Three Dimensional Geometry on Quantrex Academy →

More from Three Dimensional Geometry

Let P be the plane such that it contains the straight line (x-1)/(2) = (y-3)/(3) =… JEE Advanced 2026Let L be the straight line joining the points P(1, 2, -1) and Q(2, 3, 1) . Let S be the… JEE Advanced 2026Let the foot of perpendicular from the point (λ, 2, 3) on the line (x-4)/(1) = (y-9)/(2)… JEE Main 2026Let a line L_1 pass through the origin and be perpendicular to the lines L_2: r = (3+t) i… JEE Main 2026The shortest distance between the lines (x-4)/(1) = (y-3)/(2) = (z-2)/(-3) and (x+2)/(2)… JEE Main 2026Let the image of the point P(0, -5, 0) in the line (x-1)/(2) = (y)/(1) = (z+1)/(-2) be… JEE Main 2026Let the image of the point P(1, 6, a) in the line L: (x)/(1) = (y-1)/(2) = (z-a+1)/(b)… JEE Main 2026Let a line L be perpendicular to both the lines L_1: (x+1)/(3) = (y+3)/(5) = (z+5)/(7)… JEE Main 2026Let a triangle PQR be such that P and Q lie on the line (x+3)/(8) = (y-4)/(2) = (z+1)/(2)… JEE Main 2026If the distance of the point (a, 2, 5) from the image of the point (1, 2, 7) in the line… JEE Main 2026The square of the distance of the point P(5, 6, 7) from the line (x-2)/(2) = (y-5)/(3) =… JEE Main 2026The square of the distance of the point of intersection of the lines r = ( i + j - k ) +… JEE Main 2026The shortest distance between the lines r =((1)/(3) i +2 j +(8)/(3) k )+λ(2 i -5 j +6 k )… JEE Main 2026If (2α+1, α^2-3α, (α-1)/(2)) is the image of (α, 2α, 1) in the line… JEE Main 2026A line with direction ratios 1, -1, 2 intersects the lines (x)/(2) = (y)/(3) = (z+1)/(3)… JEE Main 2026The square of the distance of the point (-2, -8, 6) from the line (x-1)/(1) = (y-1)/(2) =… JEE Main 2026 Full Three Dimensional Geometry list Three Dimensional Geometry JEE Maths PYQsJEE Mathematics PYQs All JEE Main PYQs