Two vectors have magnitudes 3 unit and 4 unit respectively.

$|a|=3$ and $|b|=4$

Let $\theta$ be the angle between them.

Then, using the relation $R^{2}=A^{2}+B^{2}+2 A B \cos \theta$

(a)

We get for $R=1$,

$1=9+16+24 \operatorname{Cos} \theta$

Or, $\theta=180^{\circ}$

(b)

For, $\mathrm{R}=5$, we have

$25=9+16+24 \operatorname{Cos} \theta$

Or, $\cos \theta=0$;

$\theta=90^{\circ}$

(c) For $\mathrm{R}=7$,

$49=9+16+24 \operatorname{Cos} \theta$,

Or $\cos \theta=1$,

And $\theta=0^{\circ}$