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# Solution for Find the Angle Between the Vectors $\Vec{A} = 2 \Hat{I} - 3 \Hat{J} + \Hat{K} \Text{ and } \Vec{B} = \Hat{I} + \Hat{J} - 2 \Hat{K}$ - CBSE (Commerce) Class 12 - Mathematics

ConceptBasic Concepts of Vector Algebra

#### Question

Find the angle between the vectors $\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$

#### Solution

$\text{ Let }\theta \text{ be the angle between } \vec{a} \text{ and } \vec{b} .$
$\left| \vec{a} \right| = \sqrt{\left( 2 \right)^2 + \left( - 3 \right)^2 + \left( 1 \right)^2} = \sqrt{14}$
$\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( - 2 \right)^2} = \sqrt{6}$
$\vec{a} . \vec{b} = 2 - 3 - 2 = - 3$
$\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{- 3}{\sqrt{14}\sqrt{6}} = \frac{- 3}{\sqrt{84}}$
$\Rightarrow \theta = \cos^{- 1} \left( \frac{- 3}{\sqrt{84}} \right)$

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Solution Find the Angle Between the Vectors $\Vec{A} = 2 \Hat{I} - 3 \Hat{J} + \Hat{K} \Text{ and } \Vec{B} = \Hat{I} + \Hat{J} - 2 \Hat{K}$ Concept: Basic Concepts of Vector Algebra.
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