Let a=2^i+^j−2^k, b=^i+^j and c be a vector such that |c−a|=3,|(a×b)×c|=3 and the angle between c and a×b is 30∘. Then, a⋅c is equal to
2
We have,
a=2^i+^j−2^k⇒|a|=√4+1+4=3and b=^i+^j⇒|b|=√1+1=√2Now, |c−a|=3⇒|c−a|2=9⇒(c−a).(c−a)=9⇒|c|2+|a|2−2c.a=9Again, |(a×b)×c|=3⇒|a×b||c| sin 30o=3⇒|c|=6|a×b|But a×b=∣∣
∣
∣∣^i^j^k21−2110∣∣
∣
∣∣=2^i−2^j+^k∴ |c|=6√4+4+1=2
FromEqs. (i) and (ii), we get
⇒ (2)2+(3)2−2c.a=9⇒ c.a=2