how is option "a" the answer?
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As we know
A×B=|A|. |B| sin@
where angle b/w A&B is @
&B×A=|A|. |B| sin(π-@)
For equating. LHS=RHS
we need a zero value result of sin or
sin value which has value such that sin@=sin(π-@) ..
here only sinπ. has zero value which say sfy the given condition
A×B=|A|. |B| sin@
where angle b/w A&B is @
&B×A=|A|. |B| sin(π-@)
For equating. LHS=RHS
we need a zero value result of sin or
sin value which has value such that sin@=sin(π-@) ..
here only sinπ. has zero value which say sfy the given condition
rubykermani:
Thank you so much!
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