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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. enter image description here

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?

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1 Answer 1

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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)

Diagram :)

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

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

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