Answer
See below
Work Step by Step
Given $\{v_1,v_2,v_3\}$ is linearly independent and $v_4$ is not in span $\{v_1,v_2,v_3\}$
Let $a,b,c,d$ be scalars
Obtain $av_1+bv_2+cv_3+dv_4=0$
If we let $d \ne 0 \rightarrow v_4=-\frac{a}{d}v_1-\frac{b}{d}v_2-\frac{c}{d}v_3\\
\rightarrow v_4 \in span \{v_1,v_2,v_3\}$
If $d =0 \rightarrow av_1+bv_2+cv_3+dv_4\\
=av_1+bv_2+cv_3+0v_4\\
=av_1+bv_2+cv_4\\
=0$
Since $\{v_1,v_2,v_3\}$ is a linearly independent in $V$, then $a=b=c=0$
Consequently, with $a=b=c=d=0$, the linear combination of the set of vectors $v_1,v_2,v_3,v_4$ is trivial. Therefore, $\{v_1,v_2,v_3,v_4\}$ is a linearly independent set in $V$.
