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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 - CBSE (Arts) Class 12 - Mathematics

ConceptBasic Concepts of Vector Algebra

Question

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

Solution

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

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Solution Find the Angle Between the Vectors → a and → B → a = 3 ^ I − 2 ^ J − 6 ^ K and → B = 4 ^ I − ^ J + 8 ^ K Concept: Basic Concepts of Vector Algebra.
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