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Find the Angle Between the Vectors → a and → B → a = 3 ^ I − 2 ^ J − 6 ^ K and → B = 4 ^ I − ^ J + 8 ^ K - Mathematics

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प्रश्न

Find the angle between the vectors \[\vec{a} \text{ and } \vec{b}\] \[\vec{a} = 3\hat{i} - 2\hat{j} - 6\hat{k} \text{ and } \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}\]

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उत्तर

\[\text { Let }\theta \text{ be }\ \text{ the angle between } \vec{a} \text{ and } \vec{b} . \]
\[\left| \vec{a} \right| = \sqrt{\left( 3 \right)^2 + \left( - 2 \right)^2 + \left( - 6 \right)^2} = \sqrt{49} = 7\]
\[\left| \vec{b} \right| = \sqrt{\left( 4 \right)^2 + \left( - 1 \right)^2 + \left( 8 \right)^2} = \sqrt{81} = 9\]
\[ \vec{a} . \vec{b} = 12 + 2 - 48 = -34 \]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{-34}{\left( 7 \right)\left( 9 \right)} = \frac{-34}{63}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{-34}{63} \right)\]

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अध्याय 24: Scalar Or Dot Product - Exercise 24.1 [पृष्ठ ३०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 24 Scalar Or Dot Product
Exercise 24.1 | Q 5.2 | पृष्ठ ३०

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