JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \((\alpha, \beta, \gamma)\) be the point \((8,5,7)\) in the line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}\). Then \(\alpha+\beta+\gamma\) is equal to
- A \(16\)
- B \(18\)
- C \(14\)
- D \(20\)
Answer & Solution
Correct Answer
(C) \(14\)
Step-by-step Solution
Detailed explanation
\( \overrightarrow{\mathrm{AM}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})=0 \) \( (2 \lambda-7)(2)+(3 \lambda-6)(3)+(5 \lambda-5)(5)=0 \) \( 38 \lambda=57 \) \( \lambda=\frac{3}{2} \) \( \mathrm{M}\left(4, \frac{7}{2}, \frac{19}{2}\right) \)…
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