Answer
(a) ${u} \cdot {v}=0$.
(b) ${v} \cdot {v}=14$.
(c) $\|u\|^{2}=6$.
(d) $({u} \cdot {v})v=\left[\begin{array}{l}{0}\\ {0} \\ {0}\end{array}\right]$.
(e) ${u} \cdot {(5v)}=0$.
Work Step by Step
Given ${u}=\left[\begin{array}{l}{-1} \\ {1}\\ {-2}\end{array}\right]$ and ${v}=\left[\begin{array}{l}{1}\\ {-3} \\ {-2}\end{array}\right]$, then we have
(a) ${u} \cdot {v}={u}^{T} {v}=[-1 \ \ 1 \ \ -2 ]\left[\begin{array}{1} {1} \\ {-3} \\ {-2}\end{array}\right]=[(1)(-1)+(1)(-3)+(-2)(-2)]=0$.
(b) ${v} \cdot {v}={v}^{T} {v}=[1 \ \ -3 \ \ -2 ]\left[\begin{array}{1} {1} \\ {-3} \\ {-2}\end{array}\right]=[(1)(1)+(-3)(-3)+(-2)(-2)]=14$.
(c) $\|u\|^{2}={u} \cdot {u}={u}^{T} {u}=[[-1 \ \ 1 \ \ -2 ]\left[\begin{array}{l}{-1} \\ {1}\\ {-2}\end{array}\right]=[(-1)(-1)+(1)(1)+(-2)(-2)]=6$.
(d) $({u} \cdot {v})v=({u}^{T} {v}) v=0\left[\begin{array}{l}{1}\\ {-3} \\ {-2}\end{array}\right]=\left[\begin{array}{l}{0}\\ {0} \\ {0}\end{array}\right]$.
(e) ${u} \cdot {(5v)}={u}^{T} {(5v)}=[-1 \ \ 1 \ \ -2 ]\left[\begin{array}{1} {5} \\ {-15} \\ {-10}\end{array}\right]=[(-1)(5)+(1)(-15)+(-2)(-10)]=0$.
