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

मान लीजिए कि \(\vec{a}\) और \(\vec{b}\) समान परिमाण वाले सदिश हैं इस प्रकार हैं कि \(\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1\)। तो \(\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}\) = ___

  1. A \(2+4 \sqrt{2}\)
  2. B \(1+\sqrt{2}\)
  3. C \(2+\sqrt{2}\)
  4. D \(4+2 \sqrt{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2+\sqrt{2}\)

Step-by-step Solution

Detailed explanation

\(\frac{|\bar{a}+\bar{b}|+|\bar{a}-\bar{b}|}{|\bar{a}+\bar{b}|-|\bar{a}-\bar{b}|}=\sqrt{2}+1\) Apply componendo and dividendo…
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