Mathematics · 3D Geometry

JEE Main 2025 — 24 January, Morning Shift — Question 8

Let in a $\triangle A B C$ , the length of the side $A C$ be 6 , the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ . Then the area (in sq. units) of $\triangle A B C$ is

Option A:
42
Option B:
21
Correct
Option C:
56
Option D:
17

Answer: B

Step-by-step solution

Let $\mathrm{M}(3 \lambda+6,2 \lambda+7,-2 \lambda+7)$

\overrightarrow{\mathrm{BM}}=(3 \lambda+5) \hat{\mathrm{i}}+(2 \lambda+5) \hat{\mathrm{j}}+(-2 \lambda+4) \hat{\mathrm{k}}

$\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BM}}=0=3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4)$

$\rightarrow\lambda=-1$

$\overrightarrow{\mathrm{BM}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$

$|\overrightarrow{\mathrm{BM}}|=7$ Area $=\frac{1}{2} \times 6 \times 7=21$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Straight Lines in 3D Geometry

Let the line passing through the points (-1,2,1) and parallel to the line frac x-12=frac y+13=frac z4 intersect the line x+2/3=y-3/2=z-4/1…

Answer: B

Step-by-step solution

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