मराठी

If ˆ a and ˆ b are unit vectors inclined at an angle θ, prove that

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प्रश्न

 If  \[\hat{ a  } \text{ and } \hat{b }\] are unit vectors inclined at an angle θ, prove that

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

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उत्तर

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

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

\[\text{ We have }\] 

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\text{ Now },\]

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

 

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

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 23 Scalar Or Dot Product
Exercise 24.1 | Q 8.2 | पृष्ठ ३०

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