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प्रश्न
If two vectors \[\vec{a} \text{ and } \vec{b}\] are such that \[\left| \vec{a} \right| = 2, \left| \vec{b} \right| = 1 \text{ and } \vec{a} \cdot \vec{b} = 1,\] then find the value of \[\left( 3 \vec{a} - 5 \vec{b} \right) \cdot \left( 2 \vec{a} + 7 \vec{b} \right) .\]
उत्तर
\[\text{ Given that }\]
\[\left| \vec{a} \right| = 2, \left| \vec{b} \right| = 1 \text{ and } \vec{a} . \vec{b} = 1 . . . \left( 1 \right)\]
\[\text{ Now },\]
\[\left( 3 \vec{a} - 5 \vec{b} \right) . \left( 2 \vec{a} + 7 \vec{b} \right)\]
\[ = 6 \left| \vec{a} \right|^2 + 21 \vec{a} . \vec{b} - 10 \vec{b} . \vec{a} - 35 \left| \vec{b} \right|^2 \]
\[ = 6 \left| \vec{a} \right|^2 + 21 \vec{a} . \vec{b} - 10 \vec{a} . \vec{b} - 35 \left| \vec{b} \right|^2................... (\text{ We know that } \vec{a} . \vec{b} = \vec{b} . \vec{a} )\]
\[ = 6 \left| \vec{a} \right|^2 + 11 \vec{a} . \vec{b} - 35 \left| \vec{b} \right|^2 \]
\[ = 6 \left( 2 \right)^2 + 11 \left( 1 \right) - 35 \left( 1 \right)^2............... \left[ \text{ From } (1) \right]\]
\[ = 24 + 11 - 35\]
\[ = 0\]
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