Statement-1: If $\vec a$ and $\vec b$ are unit vectors inclined at an angle θ and $\hat α$ is a unit vector bisecting the angle between them, then, $\hat α=\frac{\vec a+\vec b}{2\cos θ/2}$

Statement-2: If ΔPQR is an isosceles triangle with PQ = PR = 1, then the vector representing the bisector of angle P is $\frac{\vec{PQ}+\vec{PR}}{2}$.

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 

Let S be the mid-point of QR. Then, PS is bisector of angle P.

Let the position vectors of P, Q and R are $\vec a,\vec b$ and $\vec c$ respectively. Then, the position vectors of S is $\frac{(\vec b+\vec c)}{2}$.

$∴\vec{PS}=\frac{\vec b+\vec c}{2}-\vec as$

$⇒\vec{PS}=\frac{(\vec b-\vec a)+(\vec c-\vec a)}{2}=\frac{\vec{PQ}+\vec{PR}}{2}$

So, statement-2 is true.

The unit vector along $\vec{PS}$ is $\frac{\vec{PS}}{|\vec{PS}|}=\frac{\vec{PQ}+\vec{PR}}{|\vec{PQ}+\vec{PR}|}$

∴ Unit vector bisecting the angle between unit vectors $\vec a$ and $\vec b$ is $\frac{\vec a+\vec b}{|\vec a+\vec b|}$

Now, $|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2(\vec a.\vec b)$

$⇒|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2|\vec a||\vec b|\cos θ$

$⇒|\vec a+\vec b|^2=1+1+2\cos θ=4\cos^2\frac{θ}{2}$

$⇒|\vec a+\vec b|=2\cos \frac{θ}{2}$

∴ Unit vector bisecting the angle between $\vec a$ and $\vec b$ is $\frac{\vec a+\vec b}{2\cos \frac{θ}{2}}$

Hence, both the statements are true and statement-2 is a correct explanation of statement-1.