Let →a=2^i+^j−2^k and →b=^i+^k. If →c is a vector such that →a.→c=∣∣→c∣∣,∣∣→c−→a∣∣=2√2 and the angle between (→a×→b) and →c is 300, then ∣∣∣(→a×→b)×→c∣∣∣=
A
23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
32
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A32 ∣∣→c−→a∣∣=2√2 squaring both sides, we get ⇒∣∣→c∣∣2−2∣∣→c∣∣+9=8 ⇒∣∣→c∣∣2−2∣∣→c∣∣+1=0 ⇒(∣∣→c∣∣−1)2=0 ∴∣∣→c∣∣=1 →a×→b=∣∣
∣
∣∣^i^j^k21−2110∣∣
∣
∣∣ =2^i−2^j+^k on simplification ∴∣∣∣→a×→b∣∣∣=√4+4+1=√9=3 Now , ∣∣∣(→a×→b)×→c∣∣∣=∣∣∣→a×→b∣∣∣∣∣→c∣∣sin300 =3×1×12=32