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 $\frac{\vec{PQ}+\vec{PR}}{2}$.
Answer & explanation
Correct answer: option 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.