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

ધારોકે \(\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k}\) અને \(\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}\). મે \(\overrightarrow{\mathrm{c}}\) એવો સદીશ હોય કે જેથી \(|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}\) તથા \(\vec{a} \times \vec{b}\) અને \(\vec{c}\) વચ્ચેનો ખૂણો \(60^{\circ}\) થાય, તો \(|(\vec{a} \times \vec{b}) \times \vec{c}|=\) ...........

  1. A  \(\frac{9}{2}(6-\sqrt{6})\)
  2. B \(\frac{3}{2} \sqrt{3}\)
  3. C \(\frac{3}{2} \sqrt{6}\)
  4. D \(\frac{9}{2}(6+\sqrt{6})\)
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Answer & Solution

Correct Answer

(D) \(\frac{9}{2}(6+\sqrt{6})\)

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Detailed explanation

\( |(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \frac{\sqrt{3}}{2} \) \( |\overrightarrow{c}-\overrightarrow{a}|=2 \sqrt{2} \)…
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