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JEE Advanced · Mathematics · 30. Vector Algebra

Let a=2i^+j^-k^ and b=i^+2j^+k^ be two vectors. Consider a vector c=αa+βb, α, βR. If the projection of c on the vector a+b is 32, then the minimum value of c-a×b.c equals

  1. A 15
  2. B 20
  3. C 25
  4. D 18
Verified Solution

Answer & Solution

Correct Answer

(D) 18

Step-by-step Solution

Detailed explanation

a+b=3i+3ja=2i+j-k,a=6
a+b=32,b=i+2j+k,b=6
a.b=2+2-1=3
Now,
as given projection of a+b on c=32
a+ba+b.c=32
a+b.αa+βb=18
αa2+αa.b+βb2+βa.b=18
as given 6α+3α+6β+3β=18
c=αa+βb,α+β=2 …(i)
Since a, b and c are co-planar hence a×b.c=0
Minimum value of c-a×b.c= Minimum of c.c
= Minimum of c2
a×b.c=0
now c=αa+βb and α+β=2
c=α2i+j-k+2-αi+2j+k=
c=2+αi+4-αj+k2-2α
c2=6α2-2α+4
Minimum c2=18Minimum of an2+bn+c=-D4a=-b2-4ac4a
From JEE Advanced
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