NIMCET 2008 Question Paper with solutions

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Question : 30 of 120

Marks: +1, -0

Let

A=2 i+j-2 k

and

B=i+j .

If

C

is a vector such that

A ⋅ C=|C|,|C-A|=2 √{2}

and the angle between

A \times B

and

C

is

30^{∘}

then

|(A × B) × C|

is equal to

$\;{2}/{3}$
$\;{3}/{2}$
2
3

Solution:

Given,

| {C}- {A}|=2 √{2}

⇒ ∣ {C}- {A}^{2}∣=8

⇒ C^{2}+|A|^{2}-2 C ⋅ A=8( ∵ C ⋅ A=|C|=C)

⇒ {C}^{2}+9-2 {C}=8

⇒ C^{2}-2 C+1=0

\Rightarrow

(C-1)^{2}=0

C=1

⇒ |C|=1

Now,

|( {A} × {B}) × {C}|=|| {A} × {B}|| {C}{|\sin 30^{∘} { {n↖{⋀}}}|}

={1}/{2}| {A} × {B}|

....(I)

A × B=|\table i , j , k ; 2 , 1 , -2 ; 1 , 1 , 0|

= {i}(0+2)- {j}(0+2)+k(2-1)

=2 i-2 j+k

⇒ \;\; |A × B|=√{4+4+1}=3

....(ii)
From Eqs. (i) and (ii). we get

|(A × B) × C|={3}/{2}

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