a×d−b×d=0(a−b)×d=0 d=t(a−b)d=t(−2i^−j^+2k^)∣d∣=1∣t∣=31
c⋅a=0λ+μ=0μ=−λc=λ(j^−k^),∣c∣2=2λ2c⋅d^=1t(−2,−1,2)⋅λ(0,1,−1)=1λt=3−1⇒λ2=1
∣3λ d^+μc∣2=9λ2∣ d^∣2+μ2∣c∣2+6λμ( d^⋅c)=3λ2+2λ4=5
JEE Main 2025 — Mathematics Vectors & 3D Geometry
Let a=i^+j^+k^,b=3i^+2j^−k^,c=λj^+μk^ and d^ be a unit vector such that a×d^=b×d^ and c⋅d^=1, If c is perpendicular to a, then ∣3λd^+μc∣2 is equal to _______ .
Held on 3 Apr 2025 · Verified 6 Jul 2026.
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