Three vectors →a,→b,→c are such that →a×→b=4(→a×→c) and |→a|=|→b|=1 and |→c|=14. If the angle between →b and →c is π3, then →b is
A
→a+4→c
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B
→a−4→c
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C
4→c−→a
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D
2→c−→a
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Solution
The correct options are A→a+4→c C4→c−→a →a×→b=4(→a×→c) ⇒→a×→b=(→a×4→c) ⇒→a×→b−(→a×4→c)=0 ⇒→a×(→b−4→c)=0 So, →a and (→b−4→c) are collinear or parallel vectors. ⇒λ→a=(→b−4→c)…[1] ⇒|λ→a|2=|→b−4→c|2 ⇒λ2|→a|2=|→b|2+16|→c|2−8×(→b⋅→c) ⇒λ=±1