Let c=xi^+yj^+zk^.
From a×c=b: z−y=4, z+2x=7, y+2x=3.
From c⋅(i^+j^+k^)=4: x+y+z=4.
Solving: y=−1, z=3, x=2.
c=2i^−j^+3k^.
a+c=(−1+2)i^+(2−1)j^+(2+3)k^=i^+j^+5k^.
∣a+c∣2=1+1+25=27.
JEE Main 2026 — Mathematics Vectors & 3D Geometry
Let a=−i^+2j^+2k^, b=8i^+7j^−3k^ and c be vector such that a×c=b. If c⋅(i^+j^+k^)=4, then ∣a+c∣2 is equal to :
Held on 21 Jan 2026 · Verified 6 Jul 2026.
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