If {e}_{1}=(1,1,1) \& {e}_{2}=(1,1,-1) and {a} and {b} are two vectors such that {e}_{1}=2 {{a}}^{\to}+\overrightarrow{{b}} \& {e}_{2}={{a}}^{\to}+2 {{b}}^{\to}. Then the angle between \overrightarrow{{a}} \& \overrightarrow{ b} is

Correct Answer is : cos−1(−711)

NIMCET 2021 Question Paper with solutions

Question : 42 of 120

Marks: +1, -0

{e}_{1}=(1,1,1) \ \& \ {e}_{2}=(1,1,-1)

and

{a}

and

{b}

are two vectors such that

{e}_{1}=2 {{a}}↖{→}+{{b}↖{→}}\ \&\ {e}_{2}={{a}}↖{→}+2 {{b}}↖{→}

. Then the angle between

{{a}↖{→}} \ \& \ { b}↖{→}

Solution:

e_{1}=(1,1,1)

{e}_{1}=2 {{a}↖{ →}}+{{b}}↖{ →}

3 {{a}}↖{→}=2 {e}_{2}-{e}_{1}

{a↖{ →}}={1}/{3}(2 {e}_{2}-{e}_{1})

{a}↖{ →}=\{{1}/{3}, {1}/{3},-1\}

e_{2}=(1,1,-1)

{e}_{2}={{a}↖{ →}}+2{{b}}↖{ →}

3{{b}}↖{ →}=2 {e}_{1}-{e}_{2}

{{b}}↖{ →}={1}/{3}(2 {e}_{1}-{e}_{2})

{{b}↖{ →}}=\{{1}/{3}, {1}/{3}, 1\}

\cos \θ={{a} ↖{ →}.{b}↖{ →}}/{|{a}↖{ →}||{b}↖{ →}|}

={{1}/{9}+{1}/{9}-1}/{\√{{1}/{9}+{1}/{9}+1} \√{{1}/{9}+{1}/{9}+1}}

\cos \θ={-7}/{11}

\θ=\cos ^{-1}({-7}/{11})

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