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  • Explain solution RD Sharma class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer question 16

Explain solution RD Sharma class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer question 16

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ANSWER:  2

GIVEN:   \frac{|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}+}=?

HINTS:Use algebraic identities

SOLUTION:

\frac{|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}+}= \frac{|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{a}|^{2}+\left.|\vec{b}|\right|^{2}-2 \vec{a} \cdot \vec{b}}{{|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}}}           \left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b \quad \&(a-b)^{2}=a^{2}+b^{2}-2 a b\right]

\Rightarrow \frac{|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}}=\frac{2|\vec{a}|^{2}+2|\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}}

\Rightarrow \frac{|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}}{|\vec{a}|^{2}+|\vec{b}|^{2}}=2

 

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