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If either vector \(\overrightarrow a = \overrightarrow 0\) or \(\overrightarrow b = \overrightarrow 0\), then \(\overrightarrow a . \overrightarrow b = 0\). But the converse need not be true. Justify your answer with an example.

1 Answer

  • $\overrightarrow a.\overrightarrow b=[a_1\hat i+a_2\hat j+a_3\hat k].[b_1\hat i+b_2\hat j+b_3\hat k]=a_1b_1+a_2b_2+a_3b_3$
  • $\mid\overrightarrow a\mid=\sqrt{a_1^2+a_2^2+a_3^2}$
Step 1:
Let $\overrightarrow a=\hat i-2\hat j+\hat k$ and $\overrightarrow b=\hat i+3\hat j+5\hat k$
Magnitude of $\overrightarrow a$ is $\mid\overrightarrow a\mid$
$\mid\overrightarrow a\mid=\sqrt{1^2+(-2)^2+1^2}=\sqrt 6$
Step 2:
Magnitude of $\overrightarrow b$ is $|\overrightarrow b\mid$
$\mid\overrightarrow b\mid=\sqrt{1^2+3^2+5^2}$
Step 3:
$\mid\overrightarrow a\mid=\sqrt 6$ and $\mid\overrightarrow b\mid=\sqrt {35}$
$\Rightarrow \overrightarrow a\neq 0$ and $\overrightarrow b\neq 0$
Step 4:
$\overrightarrow a.\overrightarrow b=(\hat i-2\hat j+\hat k).(\hat i+3\hat j+5\hat k)$
$\quad\quad\;=1\times 1-2\times 3+1\times 5$
$\Rightarrow \overrightarrow a.\overrightarrow b=0$,even though $\overrightarrow a\neq 0$ and $\overrightarrow b\neq 0$
answered May 21, 2013 by sreemathi.v

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