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  4. If in a vectorial triangle ABC a =2, b =3, c =4, then acute angle between veca and vecb=....

Q. If in a vectorial triangle $ABC\, a =2, b =3, c =4$, then acute angle between $\vec{a}$ and $\vec{b}=$....

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Solution:

$\vec{a}+\vec{b}+\vec{c}=\vec{o}$
$\Rightarrow \vec{a}+\vec{b}=-\vec{c}$
$\Rightarrow a^{2}+b^{2}+2 \vec{a} \cdot \vec{b}=c^{2}$
$\Rightarrow 4+9+2(2.3) \cos \theta=16$
$\therefore \cos \theta=\frac{16-13}{12}=\frac{3}{12}=\frac{1}{4}$

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