- Home
- Standard 11
- Physics
3-1.Vectors
medium
If $P + Q = R$ and $| P |=| Q |=\sqrt{3}$ and $| R |=3$, then the angle between $P$ and $Q$ is
A$\pi / 4$
B$\pi / 6$
C$\pi / 3$
D$\pi / 2$
Solution
(c)
$3=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2+2(\sqrt{3})(\sqrt{3}) \cos \theta}$
$\Rightarrow \theta=60^{\circ}$
$3=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2+2(\sqrt{3})(\sqrt{3}) \cos \theta}$
$\Rightarrow \theta=60^{\circ}$
Standard 11
Physics