Question

# Let three vectors →a,→b and →c be such that →a×→b=→c, →b×→c=→a and |→a|=2. Then which one of the following is not true:

A
a×((b+c)×(bc))=0
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B
Projection of a on (b×c) is 2
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C
[a b c]+[c a b]=8
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D
|3a+b2c|2=51
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Solution

## The correct option is D |3→a+→b−2→c|2=51∵ →a×→b=→c⇒(→a×→b).→c=→c.→c ∴[→a →b →c]=|→c|2⋯(i) and (→b×→c)=→a⇒→a.(→b×→c)=→a.→a ∴[→a →b →c]=|→a|2=4⋯(ii) ∴|→a|=|→c|=2 Option: (1) →a×((→b+→c)×(→b−→c))=→a×(−→b×→c+→c×→b) =2→a×(→c×→b)=2→a×(−→a)=→0 Option: (2) Projection of →a on (→b×→c)=→a.(→b×→c)|→b×→c|=|→a|2|→a|=2 Option: (3) [→a →b →c]+[→c →a →b]=2[→a →b →c]=8 Option: (4) |3→a+→b−2→c|2 =9→a2+→b2+4→c2+6→a.→b−4→b.→c−12→a.→c =9.22+12+4.22+0 =53 ∵ →a.→b=→b.→c=→c.→a=0 and [→a →b →c]2=∣∣ ∣ ∣ ∣∣→a.→a→a.→b→a.→c→b.→a→b.→b→b.→c→c.→a→c.→b→c.→c∣∣ ∣ ∣ ∣∣ ∴ 16=|→a|2 |→b|2 |→c|2 ∴ |→b|=1

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