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Find | → a | a N D ∣ ∣ → B ∣ ∣ If ( → a + → B ) ⋅ ( → a − → B ) = 3 a N D | → a | = 2 ∣ ∣ → B ∣ ∣ - Mathematics

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प्रश्न

Find \[\left| \vec{a} \right| and \left| \vec{b} \right|\] if 

\[\left( \vec{a} + \vec{b} \right) \cdot \left( \vec{a} - \vec{b} \right) = 3\text{  and } \left| \vec{a} \right| = 2\left| \vec{b} \right|\]

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उत्तर

\[\ \text{ Given that }\]

\[\left| \vec{a} \right| = 2 \left| \vec{b} \right| . . . \left( 1 \right)\]

\[\text{ And } \left( \vec{a} + \vec{b} \right) . \left( \vec{a} - \vec{b} \right) = 3\]

\[ \Rightarrow \left| \vec{a} \right|^2 - \left| \vec{b} \right|^2 = 3\]

\[ \Rightarrow \left( 2 \left| \vec{b} \right| \right)^2 - \left| \vec{b} \right|^2 = 3........... \left[ \text{ From } (1) \right]\]

\[ \Rightarrow 4 \left| \vec{b} \right|^2 - \left| \vec{b} \right|^2 = 3\]

\[ \Rightarrow 3 \left| \vec{b} \right|^2 = 3\]

\[ \Rightarrow \left| \vec{b} \right|^2 = 1\]

\[ \Rightarrow \left| \vec{b} \right| = 1\]

\[\left| \vec{a} \right| = 2 \left| \vec{b} \right| = 2 \left( 1 \right) = 2\]

\[ \therefore \left| \vec{a} \right| = 2 \text{ and } \left| \vec{b} \right| = 1\]

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अध्याय 24: Scalar Or Dot Product - Exercise 24.1 [पृष्ठ ३१]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 24 Scalar Or Dot Product
Exercise 24.1 | Q 31.3 | पृष्ठ ३१

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