The simple sum of two co-initial vectors is 16 units. Their vector sum is 8 units. The resultant of the vectors is perpendicular to the smaller vector. The magnitudes of the two vectors are

According to question, P + Q = 16…………(i)

$\Rightarrow P^2+Q^2+2PQ\cos \theta =64$

$\therefore \tan {90}^0=\frac{Q\sin \theta }{P+Q\cos \theta }\mathrm{............}\left(ii\right)$

or $\infty =\frac{Q\sin \theta }{P+Q\cos \theta }$

$\Rightarrow P+Q\cos \theta =0$

or $Q\cos \theta =-P\mathrm{...........}\left(iii\right)$

From Eqs (ii) and (i), we get

$P^2+Q^2+2P(-P)=64$

or $Q^2-P^2=64$

or $(Q-P)(Q+P)=\mathrm{64.............}\left(iv\right)$

Now from Eqs (i) and (iv), we get

or $Q-P=\frac{64}{16}=\mathrm{4............}\left(v\right)$

Adding, Eqs (i) and (v), we get

$2Q=20$

or $Q=10units$

From Eq(i), $P+10=16$

or $P=6units$