TS EAMCET · Maths · Three Dimensional Geometry
Let \(\mathrm{A}(\alpha, 4,7)\) and \(\mathrm{B}(3, \beta, 8)\) be two points in space. If YZ plane and ZX plane respectively divide the line segment joining the points A and B in the ratio \(2: 3\) and 4:5, then the point C which divides \(\overline{\mathrm{AB}}\) in the ratio \(\alpha: \beta\) externally is
- A \(\left(\frac{16}{3}, 10,3\right)\)
- B \(\left(\frac{-16}{3}, \frac{28}{3}, \frac{19}{3}\right)\)
- C \(\left(\frac{-16}{3}, \frac{-28}{3}, \frac{-19}{3}\right)\)
- D \(\left(\frac{-16}{3}, 10, \frac{19}{3}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{-16}{3}, 10, \frac{19}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{2(3) + 3(\alpha)}{2+3} = 0 \Rightarrow 6+3\alpha = 0 \Rightarrow \alpha = -2\) \(\frac{4(\beta) + 5(4)}{4+5} = 0 \Rightarrow 4\beta+20 = 0 \Rightarrow \beta = -5\) C divides \(\overline{\mathrm{AB}}\) in the ratio \(\alpha : \beta = -2 : -5\) externally.…
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