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NDA Math Vector Algebra

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Consider the following for the next two (02) items that follow:

Let

\vec{a},\vec{b}

and

\vec{c}

be three vectors such that

\vec{a}++\vec{b}+\vec{c}=0

and

|\vec{a}|=10,|\vec{b}|=6

and

|\vec{c}|=14

[NDA II 2016]

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Question : 64 of 101

Marks: +1, -0

What is the angle between

\vec{a}

\vec{b}

$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$75^{\circ}$

Solution:

Since

\vec{a}+\vec{b}+\vec{c}=0

\Rightarrow \vec{a}+\vec{b}=-\vec{c}

\Rightarrow |\vec{a}+\vec{b}|=|-\vec{c}|=|\vec{c}|

\Rightarrow |\vec{a}+\vec{b}|^2+2(\vec{a}\cdot\vec{b})=|\vec{c}|^2

\Rightarrow (10)^2+(6)^2+2(\vec{a}\cdot\vec{b})=(14)^2

\Rightarrow 2(\vec{a}\cdot\vec{b})=60

\Rightarrow \vec{a}\cdot\vec{b}=30

\Rightarrow \cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}=\frac{30}{10\times6}=\frac{1}{2}

\Rightarrow \theta=\cos^{-1}(1/2)\Rightarrow \theta=60^{\circ}

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