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Let $|\bar{a}|=7,|\bar{b}|=11,|\bar{a}+\bar{b}|=10 \sqrt{3}$
What is the angle between $(\bar{a}+\bar{b})$ and $(\bar{a}-\bar{b})$ ?
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What is the angle between $(\bar{a}+\bar{b})$ and $(\bar{a}-\bar{b})$ ?
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The correct answer is:
None of these
Let angle between $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$ be $\alpha$
$\cos \alpha=\frac{(\vec{a}+\vec{b})(\vec{a}-\vec{b})}{|\vec{a}+\vec{b} \| \vec{a}-\vec{b}|}$
$=\frac{(7)^{2}-(11)^{2}}{10 \sqrt{3} \times 2 \sqrt{10}}=\frac{(7+11)(7-11)}{20 \sqrt{3} \times \sqrt{10}}=\frac{-18}{5 \sqrt{30}}$
$=\frac{-6 \times 3}{5 \sqrt{30}} \times \frac{\sqrt{30}}{\sqrt{30}}=-\frac{3 \sqrt{30}}{25}$
$\alpha=\cos ^{-1}\left(\frac{-3}{5} \sqrt{\frac{6}{5}}\right)$
$\cos \alpha=\frac{(\vec{a}+\vec{b})(\vec{a}-\vec{b})}{|\vec{a}+\vec{b} \| \vec{a}-\vec{b}|}$
$=\frac{(7)^{2}-(11)^{2}}{10 \sqrt{3} \times 2 \sqrt{10}}=\frac{(7+11)(7-11)}{20 \sqrt{3} \times \sqrt{10}}=\frac{-18}{5 \sqrt{30}}$
$=\frac{-6 \times 3}{5 \sqrt{30}} \times \frac{\sqrt{30}}{\sqrt{30}}=-\frac{3 \sqrt{30}}{25}$
$\alpha=\cos ^{-1}\left(\frac{-3}{5} \sqrt{\frac{6}{5}}\right)$
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