We have,
∣a∣=3=a
a⋅c=∣c∣=c
Now,
∣c−a∣=22
⇒∣c−a∣2=8
⇒c2+a2−2c⋅a=8
⇒c2+9−2(c)=8
⇒c2−2c+1=0
⇒(c−1)2=0
⇒c=1=∣c∣
Now,
a×b=∣i^21j^11k^−20∣
⇒a×b=2i^−2j^+k^
⇒∣a×b∣=22+(−2)2+12=3
Now,
∣(a×b)×c∣=∣a×b∣∣c∣sin(6π)
⇒∣(a×b)×c∣=3⋅1⋅21=23
JEE Main 2021 — Mathematics Vectors & 3D Geometry
Let a=2i^+j^−2k^ and b=i^+j^. If c is a vector such that a⋅c=∣c∣,∣c−a∣=22 and the angle between (a×b) and c is 6π, then the value of ∣(a×b)×c∣ is:
Held on 20 Jul 2021 · Verified 6 Jul 2026.
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