Let ΔPQR be a triangle. Let a = QR, b = RP and c = PQ. If |a|=12, |b|=4√3 and b.c=24, then which of the following is/are true ?
|c|22−|a|=12
|a×b+c×a|=48√3
a.b=−72
Given, |a|=12, |b|=4√3|a|=12, |b|=4√3a+b+c=0⇒ a=−(b+c)
We have, |a|2=|b+c|2
⇒ |a|2=|b|2+|c|2+2b.c⇒ 144=48+|c|2+48⇒ |c|2=48⇒|c|=4√3Also, |c|2=|a|2+|b|2+2a.b⇒ 48=144+48+2 a.b⇒ a.b=−72
∴ Option (d) is correct.
Also, a×b=c×a
⇒a×b+c×a=2a×b⇒|a×b+c×a|=2a×b⇒|a×b+c×a|=2|a×b|=2√|a|2|b|2−(a.b)2=2√(144)(48)−(−72)2=2(12)√48−36=48√3
∴ Option (c) is correct.
Also,|c|22−|a|=24−12=12
∴ Option (a) is correct.
and |c|22+|a|=24+12=36
∴ Option (b) is not correct.