If a^{\to} and b^{\to} are vectors in space, given by a^{\to} = {i^{\^}-2j^{\^}}/\sqrt5 and b^{\to} = {2i^{\^}+j^{\^}+3k^{\^}}/\sqrt{14} then the value of (2a^{\to}+b^{\to}) . [(a^{\to}\timesb^{\to})\times(a^{\to}-2b^{\to})] is :

NIMCET 2015 Question Paper with solutions

Question : 22 of 120

Marks: +1, -0

a↖{→}

and

b↖{→}

are vectors in space, given by

a↖{→}

{i↖{\^}-2j↖{\^}}/√5

and

b↖{→}

{2i↖{\^}+j↖{\^}+3k↖{\^}}/√{14}

then the value of

(2a↖{→}+b↖{→})

[(a↖{→}×b↖{→})×(a↖{→}-2b↖{→})]

is :

Solution:

We have,

a↖{→}={i↖{⋀}-2j↖{⋀}}/√5

and

b={2i↖{⋀}+j↖{⋀}+3↖{⋀}}/√14

(2a+b)⋅[(a×b)×(a-2b)]

⇒(2a+b)⋅[a⋅(a-2b)b⋅(a-2b)a]

[∵(a×b)×c=(a-c)b-(b-c)a]

⇒(2a+b)⋅[(a^2-2a⋅b)b-(b⋅a-2b^2)a]

⇒[2a+b]⋅[(1-0)b-(1-2)a]

∵a^2=1=b^2,a⋅b=0

[2a+b]⋅[b+2a]

⇒4a^2+4a⋅b+b^2

⇒4+0+1=5

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