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Given that $\overrightarrow a .\overrightarrow b = 0$ and $\overrightarrow a \times \overrightarrow b = 0$. What can you conclude about the vectors $\overrightarrow a $ and $\overrightarrow b $ ?

Answer Verified Verified
Hint: Here, we need to draw a conclusion about the vectors $\overrightarrow a $ and $\overrightarrow b $from the statements $\overrightarrow a .\overrightarrow b = 0$ and $\overrightarrow a \times \overrightarrow b = 0$ by considering $\overrightarrow a .\overrightarrow b = \left| a \right|.\left| b \right|.\cos \theta $and $\overrightarrow a \times \overrightarrow b = \left| a \right|.\left| b \right|.\sin \theta $.

Complete step-by-step answer:

Given,
i. $\overrightarrow a .\overrightarrow b = 0$.
Here, $\overrightarrow a .\overrightarrow b = 0$ is the dot product of the vectors $\overrightarrow a $ and $\overrightarrow b $.As, we know the dot product of two vectors can be written as:
$\overrightarrow a .\overrightarrow b = \left| a \right|.\left| b \right|.\cos \theta \to (1)$
Where:
 $\left| a \right|$ Is the magnitude of$\overrightarrow a $, $\left| b \right|$is the magnitude of $\overrightarrow b $and $\theta $ is the angle between $\overrightarrow a $ and $\overrightarrow b $.
It is given that $\overrightarrow a .\overrightarrow b = 0$ i.e..,

$\left| a \right|.\left| b \right|.\cos \theta = 0 \to (2)$
So, from equation (2) we can say that the dot product of vectors $\overrightarrow a $ and $\overrightarrow b $is ‘0’ in the following cases.
(i) $\left| a \right| = 0$i.e.., the magnitude of $\overrightarrow a $is zero.
(ii) $\left| b \right| = 0$i.e.., the magnitude of $\overrightarrow b $is zero.
(iii) $\overrightarrow a \bot \overrightarrow b $i.e.., the angle between the vectors is${90^o}$$[\because \cos {90^o} = 0]$.
Hence, we can conclude that $\overrightarrow a .\overrightarrow b = 0$if ‘$\left| a \right| = 0$’or if ‘$\left| b \right| = 0$’or ‘if the vectors are perpendicular to each other.

ii. $\overrightarrow a \times \overrightarrow b = 0$.
Here, $\overrightarrow a \times \overrightarrow b = 0$ is the cross product of the vectors $\overrightarrow a $ and $\overrightarrow b $.As, we know the cross product of two vectors can be written as:
$\overrightarrow a \times \overrightarrow b = \left| a \right|.\left| b \right|.\sin \theta \to (1)$
Where:
 $\left| a \right|$ Is the magnitude of$\overrightarrow a $, $\left| b \right|$is the magnitude of $\overrightarrow b $and $\theta $ is the angle between $\overrightarrow a $ and $\overrightarrow b $.
It is given that $\overrightarrow a \times \overrightarrow b = 0$ i.e..,

$\left| a \right|.\left| b \right|.\sin \theta = 0 \to (2)$
So, from equation (2) we can say that the cross product of vectors $\overrightarrow a $ and $\overrightarrow b $is ‘0’ in the following cases
(i) $\left| a \right| = 0$i.e.., the magnitude of $\overrightarrow a $is zero.
(ii) $\left| b \right| = 0$i.e.., the magnitude of $\overrightarrow b $is zero.
(iii)$\overrightarrow a \parallel \overrightarrow b $i.e.., the angle between the vectors is${0^o}$$[\because \sin {0^o} = 0]$.
Hence, we can conclude that $\overrightarrow a \times \overrightarrow b = 0$if ‘$\left| a \right| = 0$’or if ‘$\left| b \right| = 0$’or ‘if the vectors are parallel to each other.

Note: The dot product of two vectors will be $'0'$ if the vectors are perpendicular to each other (in case vectors are non-zero).Similarly, the cross product of two vectors will be $'0'$ if the vectors are parallel to each other (in case vectors are non-zero).
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