Since (a+b) is perpendicular to c, hence
(a+b)⋅c=0
⇒a⋅c+b⋅c=0
⇒5+1+2+5b1+b2+2=0
⇒5b1+b2=−10...(1)
Given,
∣a∣b⋅a=∣a∣
∵p=ri^+sj^+tk^⇒∣p∣=r2+s2+t2
⇒b1+b2+2=4
⇒b1+b2=2...(2)
Solving (1) and (2), b1=−3,b2=5
∴∣b∣=9+25+2=6
JEE Main 2019 — Mathematics Vectors & 3D Geometry
Let a=i^+j^+2k^, b=b1i^+b2j^+2k^ and c=5i^+j^+2k^ be three vectors such that the projection vector of b on a is ∣a∣ . If a+b is perpendicular to c , then ∣b∣ is equal to:
Held on 9 Jan 2019 · Verified 6 Jul 2026.
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