If a^{\to},b^{\to} are vectors such that |a^{\to}+b^{\to}| = \sqrt{29} and a^{\to}\times(2i^{\^}+3j^{\^}+4k^{\^}) = (2i^{\^}+3j^{\^}+4k^{\^}) × b^{\^} then a possible value of (a^{\to}+b^{\to}) . (-7i^{\^}+2j^{\^}+3k^{\^}) is :

NIMCET 2016 Question Paper with solutions

Question : 93 of 120

Marks: +1, -0

a↖{→},b↖{→}

are vectors such that

|a↖{→}+b↖{→}|

√{29}

and

a↖{→}×(2i↖{\^}+3j↖{\^}+4k↖{\^})

(2i↖{\^}+3j↖{\^}+4k↖{\^})

b↖{\^}

then a possible value of

(a↖{→}+b↖{→})

(-7i↖{\^}+2j↖{\^}+3k↖{\^})

is :

Solution:

Given

|a+b|=√29

and

a×(2i↖{⋀}+3j↖{⋀}+4k↖{⋀})=(2i↖{⋀}+3j↖{⋀}+4k↖{⋀})×b

⇒a×(2i↖{⋀}+3j↖{⋀}+4k↖{⋀})=-b×(2i↖{⋀}+3j↖{⋀}+4k↖{⋀})

⇒(a+b)×(2i↖{⋀}+3j↖{⋀}+4k↖{⋀})=0

⇒a+b=λ(2i↖{⋀}+3j↖{⋀}+4k↖{⋀})

⇒|a+b|=|λ|√{4+9+16}

⇒√29=|λ|√29

⇒|λ|=1

⇒λ=±1

∴\;a+b=±(2i↖{⋀}+3j↖{⋀}+4k↖{⋀})

Hence

(a+b)⋅(-7i↖{⋀}+2j↖{⋀}+3k↖{⋀})

⇒(2i↖{⋀}+3j↖{⋀}+4k↖{⋀})⋅(-7i↖{⋀}+2j↖{⋀}+3k↖{⋀})

⇒-14+6+12=4

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