MCQ
જો $\vec a + \,\vec b \, + \,\,\vec c \,\, = \,\,\vec 0 ,\,|\vec a |\, = \,\,3,\,\,|\vec b |\,\, = \,\,5$ અને $\,|\vec c |\,\, = \,\,7$ તો $\vec a $ અને $\vec b $ વચ્ચેનો ખૂણો મેળવો.
- A$\pi/3$
- B$\pi/2$
- C$\pi/6$
- D$\pi/4$
$a+b+c=0$
$\Rightarrow a+b=-c$
$\Rightarrow| a + b |=|- c |$
$\Rightarrow|a+b|^2=|-c|^2$
$\Rightarrow| a |^2+| b |^2+2 a \cdot b =| c |^2$
$\Rightarrow 9+25+2 a \cdot b =49$
Also $ab =3 \times 5$
Hence $9+25+2 ab \cos \theta=49$
$\Rightarrow 15=30 \cos \theta$
Or $\cos \theta=\frac{1}{2}$
$\theta=60^{\circ}$
so angle is $60^{\circ}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.