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

વર્તુળનો એક ચાપ \(PQ\) તેના કેન્દ્ર \(O\) પર કાટકોણ આંતરે છે.ચાપ \(PQ\) નું મધ્યબિંદુ \(R\) છે.જો \(\overrightarrow{O P}=\vec{u}\), \(\overrightarrow{O R}=\vec{v}\) અને \(\overrightarrow{O Q}=\alpha \vec{u}+\beta \vec{v}\) હોય, તો \(\alpha, \beta^2\) એ \(.......\) સમીકરણના બીજ છે.

  1. A \(x ^2- x -2=0\)
  2. B \(3 x^2+2 x-1=0\)
  3. C \(x^2+x-2=0\)
  4. D \(3 x ^2-2 x -1=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(x ^2- x -2=0\)

Step-by-step Solution

Detailed explanation

\(|\vec{u}|=|\vec{v}|=|\alpha \vec{u}+\beta \vec{v}|\) \((\vec{u}) \cdot(\alpha \vec{u}+\beta \vec{v})=0\) \(\vec{u} \cdot \vec{v}=|u||v| \cos 45^{\circ}\) \(\alpha=-\frac{\beta}{\sqrt{2}}\) \(=|\alpha \vec{u}+\beta \vec{v}|=r\) \(\alpha^2+\beta^2+\sqrt{2} \alpha \beta=1\)…
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