Question

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

• A. $\cos ^{-1}\left(\frac{1}{4}\right)$
• B. $\cos ^{-1}\left(\frac{1}{3}\right)$
• C. $\cos ^{-1}\left(\frac{1}{2}\right)$
• D. $\cos ^{-1}(1)$

Answer 1

Answer: (A)

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