TS EAMCET · Maths · Vector Algebra
The position vectors of the points \(\mathrm{A}, \mathrm{B}\) are \(\vec{a}, \vec{b}\) respectively. If the position vector of the point \(C\) is \(\frac{\vec{a}}{2}+\frac{\vec{b}}{3}\), then
- A \(C\) lies inside \(\triangle \mathrm{OAB}\)
- B \(\mathrm{C}\) lies outside \(\triangle \mathrm{OAB}\) but inside \(\angle \mathrm{AOB}\)
- C \(\mathrm{C}\) lies outside \(\triangle \mathrm{OAB}\) but inside \(\angle \mathrm{OAB}\)
- D \(\mathrm{C}\) lies outside \(\triangle \mathrm{OAB}\) but inside \(\angle \mathrm{OBA}\)
Answer & Solution
Correct Answer
(A) \(C\) lies inside \(\triangle \mathrm{OAB}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{O C}=\frac{\vec{a}}{2}+\frac{\vec{b}}{3}\) Let \(D\) is mid point of \(O A\), \(\overrightarrow{O D}=\frac{\vec{a}}{2}\) Let \(E\) is a point dividing \(O B\) in \(1: 2\) \(\therefore \quad \overrightarrow{O E}=\frac{\vec{b}}{3}\) Clearly…
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