TS EAMCET · Maths · Vector Algebra
Let \(\mathrm{ABC}\) be a triangle and \(\bar{a}, \bar{b}, \bar{c}\) be the position vectors of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) respectively. If \(\mathrm{D}\) divides \(\mathrm{BC}\) in the ratio \(2: 3\) internally and \(\mathrm{E}\) divides \(\mathrm{CA}\) in the ratio \(2: 1\) internally then the position vector of the point \(\mathrm{P}\) which divides DE in the ratio \(3: 5\) internally is
- A \(\frac{1}{8}(2 \bar{a}+3 \bar{b}+3 \bar{c})\)
- B \(\frac{1}{8}(3 \bar{a}+2 \bar{b}+3 \bar{c})\)
- C \(\frac{1}{8}(3 \bar{a}+3 \bar{b}+2 \bar{c})\)
- D \(\frac{3}{8}(\bar{a}+\bar{b}+\bar{c})\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{8}(2 \bar{a}+3 \bar{b}+3 \bar{c})\)
Step-by-step Solution
Detailed explanation
Given that \({D}\) divides \({BC}\) in the ratio \(2: 3\), E divides \({AC}\) in the ratio \(2: 1\) and \({P}\) divides \({DE}\) in the ration 3:5. According to question, \(\overrightarrow{{d}}=\frac{2 \overrightarrow{{c}}+3 \overrightarrow{{b}}}{5}\)…
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