Mathematics · 3D Geometry

JEE Main 2025 — 8 April, Evening Shift — Question 46

Let the area of the triangle formed by the lines x+x+ 2=y−1=z,x−35=y−1=z−112=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1} and

x−3=y−33=z−21\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1} be AA. Then A2A^{2} is equal to ____\_\_\_\_ .

Answer: 56

Numerical answer — enter this value.

Step-by-step solution

L1=x+21=y−11=z1=λL_{1}=\frac{x+2}{1}=\frac{y-1}{1}=\frac{z}{1}=\lambda, any point on it (λ−2,λ(\lambda-2, \lambda +1,λ)+1, \lambda)

L2=x−35=y−1=z−11=μL_{2}=\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}=\mu, any point on it (5μ+3(5 \mu+3, −μ,μ+1)-\mu, \mu+1)

L3=x−3=y−33=z−21=kL_{3}=\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}=k, any point on it ( −3k-3 k, 3k+3,k+2)3 k+3, k+2)

P≡P \equiv point of intersection of L1L_{1} and L2=(−2,1,0)L_{2}=(-2,1,0)

Q=\mathrm{Q}= point of intersection of L1L_{1} and L3=(0,3,2)L_{3}=(0,3,2)

R≡R \equiv point of intersection of L2L_{2} and L3=(3,0,1)L_{3}=(3,0,1)

PQ‾=2i^+2j^+2k^\overline{P Q}=2 \hat{i}+2 \hat{j}+2 \hat{k}

PR‾=5i^−j^+k^\overline{P R}=5 \hat{i}-\hat{j}+\hat{k}

A=12∣PQ‾×PR‾∣=56A=\frac{1}{2}|\overline{P Q} \times \overline{P R}|=\sqrt{56}

A2=56A^{2}=56

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
3D Geometry
Topic
Straight Lines in 3D Geometry
Let the area of the triangle formed by the lines x+ 2=y-1=z… | JEE Main 2025 PYQ with Solution · DhiX AI