Let C→$$=A$$→$$+B$$→ then

Correct option is 2It is possible to have |C→|<|A→| and |C→|<|B→|

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Let $\vec{C} = \vec{A} + \vec{B}$ then

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C→ is always greater than |A→|

It is possible to have |C→|<|A→| and |C→|<|B→|

C is always equal to A + B

C is never equal to A + B

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detailed solution

Correct option is B

C→=A→+B→The value of C lies between A - B and A + BLet say A→=10i^ and B→=-8i^then C→=2i^∴ |C→|<|A→| or |C→|<|B→|

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Let C→=A→+B→ then

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Let $\vec{C} = \vec{A} + \vec{B}$ then