JEE Mains · Maths · STD 12 - 11. three dimension geometry
The vertices B and C of a triangle ABC lie on the line \(\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}\). The coordinates of A and B are (1, 6, 3) and \((4,9, \alpha)\) respectively and C is at a distance of 10 units from B. The area (in sq. units) of \( \Delta ABC \) is:
- A \( 5\sqrt{13} \)
- B \( 15\sqrt{13} \)
- C \( 20\sqrt{13} \)
- D \( 10\sqrt{13} \)
Answer & Solution
Correct Answer
(A) \( 5\sqrt{13} \)
Step-by-step Solution
Detailed explanation
\(\frac{4}{1}=\frac{9-1}{2}=\frac{\alpha-2}{3} \Rightarrow \alpha=14\) \(\overrightarrow{ AD } \cdot(\hat{ i }+2 \hat{ j }+3 \hat{ k })=0\) \((\lambda-1) \hat{ i }+(2 \lambda-5) \hat{ j }+(3 \lambda-1) \hat{ k }=\overrightarrow{ AD }\)…
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