NIMCET 2017 Question Paper with solutions

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Question : 101 of 120

Marks: +1, -0

Let

a↖{→},b↖{→},c↖{→}

be three vector having magnitudes 1 , 1 and 2 , respectively. If

a↖{→}×(a↖{→}×c↖{→})-b↖{→}

= 0 , then the acute angle between

a↖{→}

and

c↖{→}

is

$π/4$
$π/6$
$π/3$
None of these

Solution:

a↖{→}×(a↖{→}×c↖{→})-b↖{→}

= 0
⇒

a↖{→}×(a↖{→}×c↖{→})

=

b↖{→}

⇒

|a↖{→}×(a↖{→}×c↖{→})|

=

|b↖{→}|

⇒

|a↖{→}||a↖{→}×c↖{→}|

sin 90° =

|b↖{→}|

⇒ 1 ×

|a↖{→}×c↖{→}|

× 1 = 1
⇒

|a↖{→}×c↖{→}|

= 1
⇒

|a↖{→}||c↖{→}|sinθ

= 1
⇒ sin θ =

1/2

⇒ θ =

π/6

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