If →a,→b and →c are vectors such that →a⋅→b=0 and →a+→b=→c, then
A
|→a|2+|→b|2=|→c|2
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B
|→b|+|→c|=|→a|
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C
|→a|2+|→c|2=|→b|2
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D
|→b|2+|→c|2=|→a|2
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Solution
The correct option is A|→a|2+|→b|2=|→c|2 Given : →a⋅→b=0 →a+→b=→c
Squaring both sides, (→a+→b)2=→c2 (→a+→b)⋅(→a+→b)=|→c|2 →a⋅→a+→a⋅→b+→b⋅→a+→b⋅→b=|→c|2 |→a|2+0+0+|→b|2=|→c|2 ∴|→a|2+|→b|2=|→c|2