NIMCET 2013 Question Paper with solutions

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Question : 2 of 120

Marks: +1, -0

Let

a↖{→}

=

j↖{→}-k↖{→}

and

c↖{→}

=

i↖{→}-j↖{→}-k↖{→}

. Then the vector

b↖{→}

satisfying

(a↖{→}×b↖{→})

+

c↖{→}

= 0 and

a↖{→}.b↖{→}

= 3 is :

$-i↖{→}+j↖{→}-2k↖{→}$
$2i↖{→}-j↖{→}+2k↖{→}$
$i↖{→}-j↖{→}-2k↖{→}$
$i↖{→}+j↖{→}-2k↖{→}$

Solution:

Given,

a⋅b=3

and

(a×b)+c=0

...(i)
Multiplying Eq. (i) by b

b⋅(a×b)+b⋅c=0

⇒b⋅c=0

Let

b=xi↖{⋀}+yj↖{⋀}+zk↖{⋀}

,we get

y-z=3

and

x-y-z=0

From these equations,

x=2z+3,y=z+3

⇒b=(2z+3)i↖{⋀}+(z+3)j↖{⋀}+zk↖{⋀}

∵(a×b)+c=0

So,

z=-2

Thus,

b=-i↖{⋀}+j↖{⋀}-2k↖{⋀}

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