NIMCET 2017 Question Paper with solutions

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Question : 99 of 120

Marks: +1, -0

If

a↖{→},b↖{→},c↖{→}

are vectors such that

a↖{→}+b↖{→}+c↖{→}

= 0 and

|a↖{→}|

= 7 ,

|b↖{→}|

= 5 ,

|c↖{→}|

= 3 , hen the angle between the vectors

b↖{→}

and

c↖{→}

is

60°
30°
45°
90°

Solution:

a↖{→}+b↖{→}+c↖{→}

= 0
⇒

b↖{→}+c↖{→}

= -

a↖{→}

⇒

|b↖{→}+c↖{→}|

=

|a↖{→}|

⇒

|b↖{→}+c↖{→}|^2

=

|a↖{→}|^2

⇒

(b↖{→}+c↖{→}).(b↖{→}+c↖{→})

=

|a↖{→}|^2

⇒

|b↖{→}|^2+|c↖{→}|^2

+

2b↖{→}.c↖{→}

=

|a↖{→}|^2

⇒ 25 + 9 +

2|b↖{→}||c↖{→}|

cosθ = 49 [where θ is angle between

b↖{→}

and

c↖{→}

]
⇒ 2 × 5× 3 cosθ = 15
⇒ cosθ =

1/2

⇒ θ = 60°

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