Answer
See below
Work Step by Step
Given:
$p_1(x)=a+bx\\
p_2(x)=c+dx$
be vectors in $P_1(R)$
1. $
=aa+bb=a^2+b^2 \gt 0$ If $p(x)=0+0x \rightarrow
=0^2+0^2=0$ If $p(x) \ne 0 \rightarrow
=a^2+b^2 \gt 0$ if $a \ne 0, a^2 \gt 0 ;b \ne 0, b^2 \gt0$ Hence, $
=0$ if and only if $p(x)=0+0x$ 2. $
=ac+bd=ca+db=$
3. With scalar $k$, we have $kp_1(x)=ka+kbx$
and $=(ka)c+(kb)d\\
=kac+kbd\\
=k(ac+bd)\\
=k