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

$\begin{array}{1 1} \text{yes,it's true} \\ \text{No, it's false} \end{array} $

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1 Answer

  • The magnitude of $|\overrightarrow a|=\sqrt{a_1^2+a_2^2+a_3^2}$
  • Where $\overrightarrow a=a_1\hat i+a_2\hat j+a_3\hat k$
Step 1:
Let $\overrightarrow a=\hat i+\hat j+\hat k$ and $\overrightarrow b=2\hat i+2\hat j+2\hat k$
First let us find the magnitude of $|\overrightarrow a|$
The magnitude of $|\overrightarrow a|=\sqrt{1^2+1^2+1^2}$
This implies $\overrightarrow a \neq 0$
Step 2:
Next let us find the magnitude of $|\overrightarrow b|$
The magnitude of $|\overrightarrow b|=\sqrt{2^2+2^2+2^2}$
This implies $\overrightarrow b \neq 0$
Step 3:
Let us determine $\overrightarrow a\times\overrightarrow b$
$\overrightarrow a\times\overrightarrow b=\begin{vmatrix}\hat i&\hat j&\hat k\\1 & 1 & 1\\2 & 2 & 2\end{vmatrix}$
Let us take 2 as the common factor from $R_3$
$\overrightarrow a\times\overrightarrow b=2\begin{vmatrix}\hat i&\hat j&\hat k\\1 & 1 & 1\\1 & 1 & 1\end{vmatrix}$
Since two rows are identical.Determinant value of $\overrightarrow a\times\overrightarrow b=0$
Hence $\overrightarrow a\times\overrightarrow b=0$
Step 4:
This proves that even if $\overrightarrow a \neq 0$ and $\overrightarrow b\neq 0$,$\overrightarrow a\times\overrightarrow b$ can be zero.
answered May 22, 2013 by sreemathi.v

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