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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let in a A B C , the length of the side A C be 6 , the vertex B be...

Let in a △ABC\triangle A B C, the length of the side ACA C be 6 , the vertex BB be (1,2,3)(1,2,3) and the vertices A,CA, C lie on the line x−63=y−72=z−7−2\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}. Then the area (in sq. units)of△ABCof \triangle \mathrm{ABC} is:

Options

  1. A1717
  2. B2121
  3. C5656
  4. D4242

Correct answer

B. 2121

Step-by-step solution

 Let M(3λ+6,2λ+7,−2λ+7)BM→=(3λ+5)i^+(2λ+5)j^+(−2λ+4)k^AC→⋅BM→=0=3(3λ+5)+2(2λ+5)−2(−2λ+4)BM→=2i^+3j^+6k^∣BM→=7 Area =12×6×7=21\begin{aligned} & \text{ 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) \\ & \overrightarrow{\mathrm{BM}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\ & \mid \overrightarrow{\mathrm{BM}}=7 \\ & \text{ Area }=\frac{1}{2} \times 6 \times 7=21\end{aligned}
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