Let a → = 2 i ^ + j ^ - k ^ and b → = i ^ + 2 j ^ + k ^ be two vectors. Consider a vector c…

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

Solution

∵a→+b→=3i+3j∵a→=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→+b→a→+b→.c→=32
a→+b→.αa→+βb→=18
αa→2+αa→.b→+βb→2+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 c→2
∵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
c→2=6α2-2α+4
Minimum c→2=18∴Minimum of an2+bn+c=-D4a=-b2-4ac4a

Asked in: JEE Advanced 2019 (Paper 2)

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