We have been given that
d×b=c×b
⇒(d−c)×b=0
⇒d−c=λb(∵d=c)
⇒d=c+λb
It is given that a⋅d=24
⇒a⋅c+λb⋅a=24
⇒6+λ(3−8+14)=24
⇒9λ=18
⇒λ=2
∴d=c+2b
⇒d=8i^−5j^+18k^
∣d∣2=64+25+324=413
Hence, this is the correct option.
JEE Main 2023 — Mathematics Vectors & 3D Geometry
Let a=i^+4j^+2k^,b=3i^−2j^+7k^ and c=2i^−j^+4k^. If a vector d satisfies d×b=c×b and d⋅a=24, then ∣d∣2 is equal to
Held on 13 Apr 2023 · Verified 6 Jul 2026.
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