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Find the Angle Between the Vectors → a = ^ I − ^ J + ^ K and → B = ^ I + ^ J − ^ K . - Mathematics

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प्रश्न

Find the angle between the vectors \[\vec{a} = \hat{i} - \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} - \hat{k} .\]

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उत्तर

\[\text{ We have }\]

\[ \vec{a} = \hat{i} - \hat{j} + \hat{k} \text{ and } \vec{b} = i + j - k \]

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

\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( - 1 \right)^2 + \left( 1 \right)^2} = \sqrt{3}\]

\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( - 1 \right)^2} = \sqrt{3}\]

\[\text{ and }\]

\[ \vec{a} . \vec{b} = 1 - 1 - 1 = - 1\]

\[\text{ Now }, \]

\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{- 1}{\sqrt{3}\sqrt{3}} = \frac{- 1}{3}\]

\[ \therefore \theta = \cos^{- 1} \left( \frac{- 1}{3} \right)\]

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अध्याय 24: Scalar Or Dot Product - very short answer [पृष्ठ ४७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 24 Scalar Or Dot Product
very short answer | Q 26 | पृष्ठ ४७

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