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

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

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Similar questions

If either a = 0 or b = 0, then a.b = 0. But the converse need not to be true. Justify your answer with an example.

\vec{a} = \vec{0} or \vec{b} = \vec{0}, then \vec{a} \times \vec{b} = \vec{0} .

\to a = \to 0

\to b = \to 0

\to a \cdot \to b = 0

If either $a = 0 or b = 0$ , then $a \times b = 0$ . Is the converse true? Justify your answer with an example.

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