# 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.....

(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?

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

• Angel between $\vec{CA}$ and $\vec{PC}$ is only $\theta$. Still, the method works. Aug 9, 2022 at 1:16
• yes im sorry. my bad Aug 9, 2022 at 1:41
• actually it is 90-$\theta$ instead of + Aug 9, 2022 at 1:44