Answer
The lengths of the vectors $x$ and $y$ are $\sqrt{21}$ and $3\sqrt{2}$ respectively.
The inner product of the vectors $x$ and $y$ is $0$.
Work Step by Step
Given,
$$x=(1,4,0,2) \qquad and \qquad y=(2,-2,1,3)$$
We know that the length of the vector $w=(a,b,c,d)$ is given as,
$$||w||=\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}$$
Therefore, the lengths of the vectors $x$ and $y$ are given as,
\begin{eqnarray}
\nonumber ||x|| &=& \sqrt{1^{2}+4^{2}+0^{2}+2^{2}}=\sqrt{1+16+4}=\sqrt{21} \\
\nonumber ||y|| &=& \sqrt{2^{2}+(-2)^{2}+1^{2}+3^{2}}=\sqrt{4+4+1+9}=\sqrt{18}
\end{eqnarray}
Therefore, the lengths of the vectors $x$ and $y$ are $\sqrt{21}$ and $3\sqrt{2}$ respectively.
The inner product of the vectors $w=(a_{1},b_{1},c_{1},d_{1})$ and $z=(a_{2},b_{2},c_{2},d_{2})$ is given as,
$$w^{T}z=a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}+d_{1}d_{2}$$
Therefore, the inner product of the vectors $x$ and $y$ is given by,
\begin{eqnarray}
\nonumber x^{T}y &=& 1(2)+4(-2)+0(1)+2(3) \\
\nonumber x^{T}y &=& 2-8+0+6=8-8 \\
\nonumber x^{T}y &=& 0
\end{eqnarray}
Therefore, the inner product of the vectors $x$ and $y$ is $0$.
