If →a,→b,→c are unit vectors such that →a⋅→b=0=→a⋅→c and the angle between →b and →c is π3, then the value of |→a×→b−→a×→c| is
A
12
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B
1
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C
2
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D
None of these
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Solution
The correct option is B1 →a,→b,→c are unit vectors such that →a⋅→b=0=→a⋅→c and the angle between →b and →c is π3 →a⋅(→b−→c)=0 i.e angle between →a and →b−→c is π2 ∴|→a×→b−→a×→c|=|→a×(→b−→c)| ⇒|→a||→b−→c|=√→b2+→c2−2→b⋅→c ⇒|→a||→b−→c|=√1+1−2(12) ∴|→a×→b−→a×→c|=1