NIMCET 2008 Question Paper with solutions

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Question : 32 of 120

Marks: +1, -0

If

A+B+C=0,|A|=3,|B|=5,|C|=7,

then the angle between

A

and

B

is

$\;{π}/{6}$
$\;{2 π}/{3}$
$\;{5 π}/{3}$
$\;{π}/{4}$

Solution:

Given,

A+B+C=0

⇒ \;\; -C=(A+B)

⇒ \;\; (-C) ⋅(-C)=(A+B) ⋅(A+B)

⇒ \;\; |C|^{2}=|A|^{2}+|B|^{2}+2(A ⋅ B)

⇒ \;\; | {C}|^{2}=| {A}|^{2}+| {B}|^{2}+2| {A}|| {B}| ⋅ \cos θ

⇒ 49=25+9+2(3 × 5 \cos θ)

where 0 is the angle between

A

and

B

.

\Rightarrow

\cos θ=\;{1}/{2}

⇒ \;\; θ=\;{π}/{3}

or

2 π-\;{π}/{3}=\;{5 π}/{3}

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