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

ધારો કે \(\hat{u}\) અને \(\hat{v}\) એકમ સદિશો છે જે એક લઘુકોણ પર નમેલા છે જેથી \(|\hat{u}\times\hat{v}|=\dfrac{\sqrt{3}}{2}\). જો \(\vec{A}=\lambda\hat{u}+\hat{v}+(\hat{u}\times\hat{v})\) હોય, તો \(\lambda\) બરાબર છે:

  1. A \(\dfrac{4}{3}(\vec{A}\cdot\hat{u})-\dfrac{2}{3}(\vec{A}\cdot\hat{v})\)
  2. B \(\dfrac{2}{3}(\vec{A}\cdot\hat{u})-\dfrac{1}{3}(\vec{A}\cdot\hat{v})\)
  3. C \(\dfrac{4}{3}(\vec{A}\cdot\hat{u})+\dfrac{2}{3}(\vec{A}\cdot\hat{v})\)
  4. D \((\vec{A}\cdot\hat{u})-\dfrac{1}{2}(\vec{A}\cdot\hat{v})\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\dfrac{4}{3}(\vec{A}\cdot\hat{u})-\dfrac{2}{3}(\vec{A}\cdot\hat{v})\)

Step-by-step Solution

Detailed explanation

આપેલ છે કે \(|\hat{u}\times\hat{v}| = \dfrac{\sqrt{3}}{2}\) અને \(\hat{u}, \hat{v}\) એકમ સદિશો છે. \(\sin\theta = \dfrac{\sqrt{3}}{2}\) કારણ કે ખૂણો \(\theta\) લઘુકોણ છે, \(\theta = \dfrac{\pi}{3}\). \(\hat{u}\cdot\hat{v} = \cos\dfrac{\pi}{3} = \dfrac{1}{2}\) આપેલ છે કે…
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