For a triangle ABC, let p=BC,q=CA and r=BA. If ∣p∣=23,∣q∣=2 and cosθ=31, where θ is the angle between p and q, then ∣p×(q−3r)∣2+3∣r∣2 is equal to:
(A)340
(B)220
(C)410
(D)200✓ Correct
Answer: (D)
Solution
In triangle ABC, BC+CA=BA, so r=p+q. Then q−3r=q−3(p+q)=−3p−2q, and p×(q−3r)=p×(−3p−2q)=−2(p×q), so ∣p×(q−3r)∣2=4∣p∣2∣q∣2sin2θ. With ∣p∣2=12,∣q∣2=4 and sin2θ=1−31=32, this equals 4⋅12⋅4⋅32=128. Also ∣r∣2=∣p+q∣2=12+4+2(23)(2)⋅31=24, so 3∣r∣2=72. The required value is 128+72=200.
JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option