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Question

2i+3j+k as a sum of two vectors out of which one is perpendicular to 2i4j+k and another is parallel to 2i4j+k is p(8i+5j+4k) + q(2i4j+k) then p+q =

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Solution

Let
b=2i4j+k
a=2i+3j+k
Let
a=a1+a2
a1=vector parallel to b
a2=vector perpendicular to b
sincea1is parallel to b
a1=λ(2i4j+k)
a1=2λi4λj+λk
also
a2=a+a1
a2=(2i+3j+k)(2λi4λj+λk)
=i(22λ)+j(3+4λ)+k(1λ)
sincea2andbareperpendicular
a2b=0
[i(22λ)+j(3+4λ)+k(1λ)](2i4j+k)=0
2(22λ)+(4)(3+4λ)+(1λ)(1)=0
44λ1216λ+1λ=0
7=21λ
λ=13
a1=13(2i4j+k)
a2=i(22(13))+j(3+4(13))+k(1(13))
=13(8i+5j+4k)
Hence
13(8i+5j+4k)13(2i4j+k)
comparingwithp(8i+5j+4k)q(2i4j+k)
p=13,q=13
Hence
p+q=1313=0

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