Let overrightarrow C = overrightarrow A + overrightarrow B then

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3-1.Vectors

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Let $\overrightarrow C = \overrightarrow A + \overrightarrow B $ then

A

$|\overrightarrow {C|} $ is always greater then $|\overrightarrow A |$

B

It is possible to have $|\overrightarrow C |\, < \,|\overrightarrow A |$ and $|\overrightarrow C |\, < \,|\overrightarrow B |$

C

$C$ is always equal to $A + B$

D

$C$ is never equal to $A + B$

Solution

(b) $\vec C + \vec A = \vec B$.

The value of $C$ lies between $A – B$ and $A + B$

$|\vec C|\; < \;|\vec A|\;\;{\rm{or}}\;\;|\vec C|\; < \;|\vec B|$

Standard 11

Physics

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