In a ΔPQR, 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:
A
∣∣→c∣∣22−∣∣→a∣∣=12
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
∣∣→c∣∣22+∣∣→a∣∣=30
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
∣∣∣→a×→b+→c×→a∣∣∣=48√3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
→a.→b=−72
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D→a.→b=−72 Using vectors summation method, −→a=→b+→c
Squaring both sides, we have: |→a|2=|→c|2+|→b|2+2→c⋅→b ⇒|→c|2=122−(4√3)2−2×24 ⇒→c=4√3 ∴|→c|22−|→a|=12
Now, we also have: −→c=→a+→b
Squaring both sides, we have: ⇒|→c|2=122+(4√3)2+2→a⋅→b ⇒144+48+2→a.→b=48⇒→a.→b=−72 ∵→a+→b+→c=→0⇒→a×→b+→a×→c=→a×→0 ⇒∣∣∣→a×→b+→c×→a∣∣∣=2∣∣∣→a×→b∣∣∣ =2√|→a|2|→b|2−(→a.→b)2=2√(144)(48)−(72)2=48√3