# A,B,C,D are four points in the space and satisfy ∣vec(AB)∣=3,∣vec(BC)∣=7,∣vec(CD)∣=11 and ∣vec(DA)∣=9. Then vec(AC).vec(BD) is I know that vec(AB)+vec(BC)+vec(CA)+vec(AD)=0 and also I observed that ∣vec(AB)∣^2+∣vec(CD)∣^2=∣vec(BC)∣^2+∣vec(DA)∣^2. But couldn't get anything from it. Any hint?

A,B,C,D are four points in the space and satisfy $\mid \stackrel{\to }{AB}\mid =3,\mid \stackrel{\to }{BC}\mid =7,\mid \stackrel{\to }{CD}\mid =11$ and $\mid \stackrel{\to }{DA}\mid =9$. Then $\stackrel{\to }{AC}.\stackrel{\to }{BD}$ is
I know that $\stackrel{\to }{AB}+\stackrel{\to }{BC}+\stackrel{\to }{CA}+\stackrel{\to }{AD}=0$ and also I observed that
$\mid \stackrel{\to }{AB}{\mid }^{2}+\mid \stackrel{\to }{CD}{\mid }^{2}=\mid \stackrel{\to }{BC}{\mid }^{2}+\mid \stackrel{\to }{DA}{\mid }^{2}$. But couldn't get anything from it. Any hint?
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$\stackrel{\to }{AC}\cdot \stackrel{\to }{BD}=\left(-\stackrel{\to }{DA}+\stackrel{\to }{DC}\right)\cdot \left(-\stackrel{\to }{DB}\right)=\stackrel{\to }{DA}\cdot \stackrel{\to }{DB}-\stackrel{\to }{DC}\cdot \stackrel{\to }{DB}=$
$=DA\cdot DB\mathrm{cos}\measuredangle ADB-DC\cdot DB\mathrm{cos}\measuredangle CDB=$
$=\frac{{9}^{2}+D{B}^{2}-{3}^{2}}{2}-\frac{{11}^{2}+D{B}^{2}-{7}^{2}}{2}=0.$
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