JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}\) and \(\vec{b}\) be the vectors of the same magnitude such that \(\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1\). Then \(\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}\) is :
- A \(2+4 \sqrt{2}\)
- B \(1+\sqrt{2}\)
- C \(2+\sqrt{2}\)
- D \(4+2 \sqrt{2}\)
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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