因 $\left|\overset{\large\rightharpoonup}{u}\right|=3$。內積 $\overset{\large\rightharpoonup}{u}\cdot\overset{\large\rightharpoonup}{v}$ 可取任意實數，故 $(1)$ 對。外積必垂直於 $\overset{\large\rightharpoonup}{u}$，但 $\left(1,3,4\right)\cdot\left(2,2,1\right)=12\ne0$，故 $(2)$ 錯。若 $\left|\overset{\large\rightharpoonup}{u}\cdot\overset{\large\rightharpoonup}{v}\right|=2\left|\overset{\large\rightharpoonup}{v}\right|$，不必平行，故 $(3)$ 錯。若 $\left|\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v}\right|=3\left|\overset{\large\rightharpoonup}{v}\right|$，則 $\sin\theta=1$，故內積為 $0$。若外積為零則兩向量平行，再加上內積為零且 $\overset{\large\rightharpoonup}{u}\ne\overset{\large\rightharpoonup}{0}$，可得 $\overset{\large\rightharpoonup}{v}=\overset{\large\rightharpoonup}{0}$。