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If ^ a and ^ B Are Unit Vectors Inclined at an Angle θ, Prove that Cos θ 2 = 1 2 ∣ ∣ ^ a + ^ B ∣ ∣

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प्रश्न

If  \[\hat{a} \text{ and } \hat{b}\] are unit vectors inclined at an angle θ, prove that \[\cos\frac{\theta}{2} = \frac{1}{2}\left| \hat{a} + \hat{b} \right|\] 

योग
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उत्तर

\[\text{ Given that } \hat{ a }\ \text{ and } \hat{b}\ \text{ are unit vectors }.\]

\[So,\left| \hat{a} \right|=1,\left| \hat{b} \right|=1\]
\[\text{We have}\]

\[ \left| \hat{a} + \hat{b} \right|^2 = \left| \hat{a} \right|^2 + \left| \hat{b} \right|^2 + 2 \hat{a} . \hat{b} \]

\[ = 1 + 1 + 2 \left| \hat{a} \right| \left| \hat{b} \right| \cos \theta\]

\[ = 2 + 2\cos \theta\]

\[ \Rightarrow \cos\theta = \frac{\left| \hat{a} + \hat{b} \right|^2 - 2}{2} .....................\left( 1 \right)\]

 

\[ \left| \hat{a} - b \right|^2 = \left| \hat{a} \right|^2 + \left| \hat{b} \right|^2 - 2 \hat{a} .\hat {b} \]

\[ = 1 + 1 - 2 \left| \hat{a} \right| \left| \hat{b} \right| \cos \theta\]

\[ = 2 - 2\cos \theta\]

\[ \Rightarrow \cos\theta = \frac{2 - \left| \hat{a} - \hat{b} \right|^2}{2}...................... \left( 2 \right)\]

\[ \text{ Now },\]

\[\cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}}\]

\[ = \sqrt{\frac{1 + \frac{\left| \hat{a} + \hat{b} \right|^2 - 2}{2}}{2}} ...............\left[\text{  From }\left( 1 \right) \right]\]

\[ = \sqrt{\frac{2 + \left| \hat{a} + \hat{b} \right|^2 - 2}{4}}\]

\[ = \sqrt{\frac{\left| \hat{a} + \hat{b} \right|^2}{4}}\]

\[ = \frac{1}{2}\left| \hat{a} + \hat{b} \right|\]

 

 

 

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अध्याय 23: Scalar Or Dot Product - Exercise 24.1 [पृष्ठ ३०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 23 Scalar Or Dot Product
Exercise 24.1 | Q 8.1 | पृष्ठ ३०

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