मराठी

Find the Angle Between Two Vectors → a and → B If | → a | = √ 3 , ∣ ∣ → B ∣ ∣ = 2 and → a ⋅ → B = √ 6

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प्रश्न

Find the angle between two vectors \[\vec{a} \text{ and } \vec{b}\] if 

\[\left| \vec{a} \right| = \sqrt{3}, \left| \vec{b} \right| = 2 \text{ and } \vec{a} \cdot \vec{b} = \sqrt{6}\] 

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उत्तर

 Let θ be the angle between \[\vec{a} \text{ and }\vec{b} .\]

Given that 

\[\left| \vec{a} \right| = \sqrt{3}, \left| \vec{b} \right| = \text{ 2 and }\vec{a} . \vec{b} = \sqrt{6} . . . \left( 1 \right)\]
We know that

\[ \vec{a} . \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta\]

\[ \Rightarrow \sqrt{6} = \left( \sqrt{3} \right)\left( 2 \right) \cos \theta .....................\left[ \text{ Using }\left( 1 \right) \right]\]

\[ \Rightarrow \cos \theta = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{1}{\sqrt{2}}\]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4}\]

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पाठ 23: Scalar Or Dot Product - Exercise 24.1 [पृष्ठ ३१]

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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 33.1 | पृष्ठ ३१
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