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# Solution for Find the Angle Between the Vectors → a = ^ I + 2 ^ J − ^ K , → B = ^ I − ^ J + ^ K - CBSE (Commerce) Class 12 - Mathematics

ConceptBasic Concepts of Vector Algebra

#### Question

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

#### Solution

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

$\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( 2 \right)^2 + \left( - 1 \right)^2} = \sqrt{6}$

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

$\vec{a} . \vec{b} = 1 - 2 - 1 = - 2$

$\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{- 2}{\sqrt{6}\sqrt{3}} = \frac{- 2}{\sqrt{18}} = \frac{- \sqrt{2} \times \sqrt{2}}{\sqrt{2} \times \sqrt{9}} = \frac{- \sqrt{2}}{3}$

$\Rightarrow \theta = \cos^{- 1} \left( \frac{- \sqrt{2}}{3} \right)$

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Solution Find the Angle Between the Vectors → a = ^ I + 2 ^ J − ^ K , → B = ^ I − ^ J + ^ K Concept: Basic Concepts of Vector Algebra.
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