the range of values of $\vec{PA}\cdot\vec{PB}$ is In $\triangle ABC,AC=3,BC=4$, $\angle C=90°$,$P$ is a moving point in the plane of triangle $ABC$ and $PC=1$,then the range of values of $\vec{PA}\cdot\vec{PB}$ is.....
Select your answer
(A) $[-5,3] $
(B) $[-3,5] $
(C) $[-6,4] $
(D)$[-4,6]​$
My attempt
I drew the triangle.

From the triangle law of vector addition, I got
$$\vec{PB }=\vec{CB}-\vec{CP}$$
$$\vec{PA}=\vec{CA}-\vec{CP}$$
$$\vec{PB} \cdot \vec{PA}=(\vec{CB}-\vec{CP})\cdot (\vec{CA}-\vec{CP})$$ $$=-||\vec{CA}||\cos(\alpha)-||\vec{CB}||\sin(\alpha)+1$$( $\because ||\vec{CP}||^2=1$)
$\alpha \in [0,\pi/2]$. right?
When $\alpha=0$, we get $\vec{PA} \cdot \vec{PA}=-2$ and When $\alpha=\pi/2$, we get $\vec{PA} \cdot \vec{PA}=-3.$ So, range of values of$\vec{PA} \cdot \vec{PA}$ is $[-3,-2]$. But none of the option matches with my option. Where is my mistake?
 A: Good Evening sir.
This is how i tackled this question.
So in this figure (sorry i don't have access to Mathematica or similar software) I have introduced a Cartesian Plane centred at $C$ so the points $A$ and $B$ lie on $(3,0)$ and $(0,4)$ respectively. The variable point P lies on the unit circle centred at origin (as per the question)

We intend to find the range of $\vec{PA}\cdot \vec{PB}\ $
First I rewrote $\vec{PA} \cdot \vec{PB}$ as $(\vec{PC}+\vec{CA})\cdot(\vec{PC}+\vec{CB})$
Distributing we get:
$$(\vec{PC}\cdot\vec{PC})+(\vec{PC}\cdot\vec{CB})+(\vec{CA}\cdot\vec{PC})+(\vec{CA}\cdot\vec{CB})$$
Now $(\vec{CA}\cdot\vec{CB})$ would be zero as $\vec{CA}\perp \vec{CB}$
Also $(\vec{PC}\cdot\vec{PC})$ would be $\lvert \vec{PC} \rvert^2$ which is $1$
Substituting we get:
$$1+(\vec{PC}\cdot\vec{CB})+(\vec{CA}\cdot\vec{PC})$$
Now I simply expand and simplify.
$$1+\lvert PC\rvert \lvert CB\rvert \cos (180-\theta) + \lvert CA\rvert \lvert PC\rvert \cos (90-\theta)$$
$$= 1+1\times 4\times (-\cos \theta)+ 3\times 1\times (\sin \theta)$$
$$= 1-4\cos \theta+3\sin \theta$$
We know that the range of $a\sin \alpha + b\cos \alpha$ is $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$
So the range of $4\cos \theta -3\sin \theta$
would be $[-5,5]$
So the range of $-4\cos \theta +3\sin \theta$
would be $[-5,5]$
So the range of $1-4\cos \theta +3\sin \theta$
would be $[-4,6]$
thus the range of $\vec{PA}\cdot \vec{PB}\ $ should be [-4,6] i.e. D option
