NIMCET 2017 Question Paper with solutions

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Question : 102 of 120

Marks: +1, -0

Let

a↖{→},b↖{→},c↖{→}

be vector such that

|a↖{→}|

= 2 ,

|b↖{→}|

= 3 ,

|c↖{→}|

= 5 and

a↖{→}+b↖{→}+c↖{→}

=

0↖{→}

. Then the value of

a↖{→}b↖{→}+b↖{→}c↖{→}+c↖{→}a↖{→}

is

38
-38
19
-19

Solution:

Now,

a↖{→}+b↖{→}+c↖{→}

= 0
⇒

a↖{→}.(a↖{→}+b↖{→}+c↖{→})

=

a↖{→}

. 0
⇒

|a↖{→}|^2+a↖{→}.b↖{→}

+

a↖{→}.c↖{→}

= 0
⇒

a↖{→}.b↖{→} + a↖{→}.c↖{→}

= - 4 ... (i)
Similarly,

b↖{→}.a↖{→}+b↖{→}.c↖{→}

= - 9 ... (ii)
and

c↖{→}.a↖{→}+c↖{→}.b↖{→}

= - 25 ... (iii)
On adding Eqs. (i), (ii) and (iii), we get
2

(a↖{→}.b↖{→}+b↖{→}.c↖{→}+c↖{→}.a↖{→})

= - 38
⇒

a↖{→}.b↖{→}+b↖{→}.c↖{→}+c↖{→}.a↖{→}

= - 19

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