JEE Mains · Maths · STD 12 - 10. vector algebra
Let PQR be a triangle. The points \(A , B\) and \(C\) are on the sides \(QR , RP\) and \(PQ\) respectively such that \(\frac{ QA }{ AR }=\frac{ RB }{ BP }=\frac{ PC }{ CQ }=\frac{1}{2}\). Then \(\frac{\operatorname{Area}(\triangle PQR )}{\operatorname{Area}(\triangle ABC )}\) is equal to \(........\)
- A \(4\)
- B \(3\)
- C \(2\)
- D \(\frac{5}{2}\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
Let \(P\) is \(\overrightarrow{0}, Q\) is \(\overrightarrow{ q }\) and \(R\) is \(\overrightarrow{ r }\) \(A\) is \(\frac{2 \overrightarrow{ q }+\overrightarrow{ r }}{3}\), B is \(\frac{2 \overrightarrow{ r }}{3}\) and \(C\) is \(\frac{\overrightarrow{ q }}{3}\) Area of…
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