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JEE Mains · Maths · STD 12 - 10. vector algebra

Let the arc \(A C\) of a circle subtend a right angle at the centre \(O\). If the point \(B\) on the arc \(A C\), divides the arc \(A C\) such that \(\frac{\text { length of } \operatorname{arc} A B}{\text { length of } \operatorname{arc} B C}=\frac{1}{5}\), and \(\overrightarrow{O C}=\alpha \overrightarrow{O A}+\beta \overrightarrow{O B}\), then \(\alpha+\sqrt{2}(\sqrt{3}-1) \beta\) is equal to

  1. A \(2 \sqrt{3}\)
  2. B \(2-\sqrt{3}\)
  3. C \(5 \sqrt{3}\)
  4. D \(2+\sqrt{3}\)
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Answer & Solution

Correct Answer

(B) \(2-\sqrt{3}\)

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Detailed explanation

\begin{aligned} & \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}} \ldots . .(1) \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\beta…

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