The Eunclidean distance between u and v is the Euclidean norm of the vector u - v. Thus, we must find

u-v=(-1,-1,8,0)-(5,6,1,4)=(-6,-7,7,-4)

and

\(\displaystyle{d}{\left({u},{v}\right)}={\left|{{\left|{{u}-{v}}\right|}}\right|}=\sqrt{{{\left(-{6}\right)}^{{2}}+{\left(-{7}\right)}^{{2}}+{7}^{{2}}+{\left(-{4}\right)}^{{2}}}}=\sqrt{{{150}}}={5}\sqrt{{{6}}}\)

Furthermore, the angle 0 between these vectors is given by

\(\displaystyle{\cos{{0}}}=\frac{{\left\langle{u},{v}\right\rangle}}{{\left|{{\left|{{u}-{v}}\right|}}\right|}}\),

where \(\displaystyle{\left\langle{u},{v}\right\rangle}\) is a scalar product of u and v. So we compute

\(\displaystyle{\left\langle{u},{v}\right\rangle}=-{5}-{6}+{8}+{0}=-{3}\)

\(\displaystyle{\left|{{\left|{{u}}\right|}}\right|}=\sqrt{{{\left(-{1}\right)}^{{2}}+{\left(-{1}\right)}^{{2}}+{8}^{{2}}+{0}^{{2}}}}=\sqrt{{{66}}}\)

\(\displaystyle{\left|{{\left|{{v}}\right|}}\right|}=\sqrt{{{5}^{{2}}+{6}^{{2}}+{1}^{{2}}+{4}^{{2}}}}=\sqrt{{{78}}}\)

Therefore,

\(\displaystyle{\cos{{0}}}=\frac{{-{3}}}{{\sqrt{{{66}}}\sqrt{{{78}}}}}\)

which means that

\(\displaystyle{0}=\frac{{\arccos{{\left(-{3}\right)}}}}{{\sqrt{{{66}}}\sqrt{{{78}}}}}\approx{1.613}\)

So, this angle is obtuse.

Result

Distance:

\(\displaystyle{d}{\left({u},{v}\right)}={5}\sqrt{{{6}}}\)

Angle:

\(\displaystyle{0}=\frac{{\arccos{{\left(-{3}\right)}}}}{{\sqrt{{{66}}}\sqrt{{{78}}}}}\approx{1.613}\)

It is obtuse.

u-v=(-1,-1,8,0)-(5,6,1,4)=(-6,-7,7,-4)

and

\(\displaystyle{d}{\left({u},{v}\right)}={\left|{{\left|{{u}-{v}}\right|}}\right|}=\sqrt{{{\left(-{6}\right)}^{{2}}+{\left(-{7}\right)}^{{2}}+{7}^{{2}}+{\left(-{4}\right)}^{{2}}}}=\sqrt{{{150}}}={5}\sqrt{{{6}}}\)

Furthermore, the angle 0 between these vectors is given by

\(\displaystyle{\cos{{0}}}=\frac{{\left\langle{u},{v}\right\rangle}}{{\left|{{\left|{{u}-{v}}\right|}}\right|}}\),

where \(\displaystyle{\left\langle{u},{v}\right\rangle}\) is a scalar product of u and v. So we compute

\(\displaystyle{\left\langle{u},{v}\right\rangle}=-{5}-{6}+{8}+{0}=-{3}\)

\(\displaystyle{\left|{{\left|{{u}}\right|}}\right|}=\sqrt{{{\left(-{1}\right)}^{{2}}+{\left(-{1}\right)}^{{2}}+{8}^{{2}}+{0}^{{2}}}}=\sqrt{{{66}}}\)

\(\displaystyle{\left|{{\left|{{v}}\right|}}\right|}=\sqrt{{{5}^{{2}}+{6}^{{2}}+{1}^{{2}}+{4}^{{2}}}}=\sqrt{{{78}}}\)

Therefore,

\(\displaystyle{\cos{{0}}}=\frac{{-{3}}}{{\sqrt{{{66}}}\sqrt{{{78}}}}}\)

which means that

\(\displaystyle{0}=\frac{{\arccos{{\left(-{3}\right)}}}}{{\sqrt{{{66}}}\sqrt{{{78}}}}}\approx{1.613}\)

So, this angle is obtuse.

Result

Distance:

\(\displaystyle{d}{\left({u},{v}\right)}={5}\sqrt{{{6}}}\)

Angle:

\(\displaystyle{0}=\frac{{\arccos{{\left(-{3}\right)}}}}{{\sqrt{{{66}}}\sqrt{{{78}}}}}\approx{1.613}\)

It is obtuse.