In a triangle ABC, if | → BC | = 3, | → CA | = 5 and | → BA | = 7,...

We have,

$\left|\overset{\to}{a}\right| = 3, \left|\overset{\to}{b}\right| = 5, \left|\overset{\to}{c}\right| = 7$

Now,

$cos\theta = \dfrac{{\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{c}\right|}^{2} - {\left|\overset{\to}{b}\right|}^{2}}{2\left|\overset{\to}{a}\right|\left|\overset{\to}{c}\right|}$

$⇒ cos\theta = \dfrac{9 + 49 - 25}{2 \cdot 3 \cdot 7}$

$⇒ cos\theta = \dfrac{11}{14}$

Projection of $\overset{\to}{c}$ on $\overset{\to}{b}$

$= | \overset{\to}{c} | cos\theta$

$= 7 \times \dfrac{11}{14}$

$= \dfrac{11}{2}$