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# Solution for If Either → a = → 0 Or → B = → 0 Then → a ⋅ → B = 0 . but the Converse Need Not Be True. Justify Your Answer with an Example. - CBSE (Commerce) Class 12 - Mathematics

ConceptBasic Concepts of Vector Algebra

#### Question

If either $\vec{a} = \vec{0} \text{ or } \vec{b} = \vec{0}$  then $\vec{a} \cdot \vec{b} = 0 .$ But the converse need not be true. Justify your answer with an example.

#### Solution

$\text{ Let us assume that either }\left| \vec{a} \right|=0 \text{ or } \left| \vec{b} \right| = 0$
$Then, \vec{a} . \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta = 0 (\theta \text{ is the angle between } \vec{a} \text{ and } \vec{b} )$
$\text{ Now, let us assume that } \vec{a} . \vec{b} = 0$
$\Rightarrow \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta = 0$
$\text{ But here we cannot say thateither }\left| \vec{a} \right|=0 or \left| \vec{b} \right| = 0 . \text{ (Because even cos } \theta \text{ can be zero) }$
$\text{ For example, let}$
$\vec{a} = 2 \hat{i} + /hat{j} + 3 \hat{k} \text{ and } \vec{b} = - 3 \hat{i} + 2 \hat{k}$
$Here,\left| \vec{a} \right|=\sqrt{4 + 1 + 9}=\sqrt{14}\neq0$
$\left| \vec{b} \right| = \sqrt{9 + 4} = \sqrt{13}\neq0$
$\text{But} \vec{a} . \vec{b} = \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) . \left( - 3 \hat{i} + 2 \hat(k) \right) = - 6 + 0 + 6 = 0$

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Solution If Either → a = → 0 Or → B = → 0 Then → a ⋅ → B = 0 . but the Converse Need Not Be True. Justify Your Answer with an Example. Concept: Basic Concepts of Vector Algebra.
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