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प्रश्न
If \[\left| \vec{a} \right| = a \text{ and } \left| \vec{b} \right| = b,\] prove that \[\left( \frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2} \right)^2 = \left( \frac{\vec{a} - \vec{b}}{ab} \right)^2 .\]
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उत्तर
\[\left( \frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2} \right)^2 \]
\[ = \left| \frac{\vec{a}}{a^2} \right|^2 + \left| \frac{\vec{b}}{b^2} \right|^2 - \frac{2 \vec{a} . \vec{b}}{a^2 b^2}\]
\[ = \frac{\left| \vec{a} \right|^2}{a^4} + \frac{\left| \vec{b} \right|^2}{b^4} - \frac{2 \vec{a} . \vec{b}}{a^2 b^2}\]
\[ = \frac{a^2}{a^4} + \frac{b^2}{b^4} - \frac{2 \vec{a} . \vec{b}}{a^2 b^2}...................(\text{ From the given information })\]
\[ = \frac{1}{a^2} + \frac{1}{b^2} - \frac{2 \vec{a} . \vec{b}}{a^2 b^2}\]
\[ = \frac{b^2 + a^2 - 2 \vec{a} . \vec{b}}{a^2 b^2}\]
\[ = \frac{a^2 + b^2 - 2 \vec{a} . \vec{b}}{a^2 b^2}\]
\[ = \frac{\left| \vec{a} \right|^2 + \left| \vec{a} \right|^2 - 2 \vec{a} . \vec{b}}{a^2 b^2}.................(\text{ From the given information })\]
\[ = \frac{\left( \vec{a} - \vec{b} \right)^2}{a^2 b^2}\]
\[ = \left( \frac{\vec{a} - \vec{b}}{ab} \right)^2\]
