1. ## vector help

AB is a diameter of a circle centred at the origin O, and P is any point on the circumference of the circle.
Using the position vectors of A, B and P, prove (using a scalar product) that AP is perpendicular to BP (i.e. the angle in the semicircle is a right angle).

Not sure where to even start. I know that the product of vectors has to equal 0 for them the be perpendicular, but I dont know the magnitude for each vector, so how can you prove they are perpendicular?

Any help appreciated.

thank you

2. Originally Posted by Tweety
Not sure where to even start. I know that the product of vectors has to equal 0 for them the be perpendicular, but I dont know the magnitude for each vector, so how can you prove they are perpendicular?

Any help appreciated.

thank you
Hi Tweety,

You may not know the lengths of $\vec{AP}$ and $\vec{BP}$

however, in using the scalar product we can work with the vectors, $\vec{OA},\ \vec{OP},\ \vec{OB}$

Since A and B are a radius away from O, then $\vec{A}$ and $\vec{B}$ are going in opposite directions,

and have the same length, hence $\vec{A}=-\vec{B}$

Then calculate the scalar product of $\vec{AP}$ and $\vec{BP}$

using $\vec{AP}=\vec{P}-\vec{A}$ and $\vec{BP}=\vec{P}-\vec{B}$

$\vec{AP}.\vec{BP}=\left(\vec{P}-\vec{A}\right).\left(\vec{P}-\vec{B}\right)$

$=|p|^2-|p||b|cos\theta-|a||p|cos({\pi}-\theta)+|a||b|cos({\pi})$

$=r^2-r^2cos\theta+r^2cos\theta-r^2$

3. Originally Posted by Tweety
Not sure where to even start. I know that the product of vectors has to equal 0 for them the be perpendicular, but I dont know the magnitude for each vector, so how can you prove they are perpendicular?

Any help appreciated.

thank you
let AB lie on the x-axis from -1 to 1

P has position vector $\vec{r_P} = \cos{t} \vec{i} + \sin{t} \vec{j}$

$AP = (\cos{t}+1) \vec{i} + \sin{t} \vec{j}$

$PB = (1-\cos{t})\vec{i} - \sin{t} \vec{j}$

$AP \cdot PB = (1-\cos^2{t}) - \sin^2{t} = 0$

4. Originally Posted by Tweety
Not sure where to even start. I know that the product of vectors has to equal 0 for them the be perpendicular, but I dont know the magnitude for each vector, so how can you prove they are perpendicular?

Any help appreciated.

thank you
First, make a sketch.

So you wan't to show that $\vec{PB}\perp \vec{PA}$, by showing that $\vec{PB}\cdot \vec{PA}=0$.

Express the two vectors as sums like this:

$\vec{PB}\cdot \vec{PA} = (\vec{PO}+\vec{OA})\cdot (\vec{PO}+\vec{OB})$
(where $O$ is the origin / center of the circle)

now use the fact that $\vec{OB}=-\vec{OA}$ to get

$=(\vec{PO}+\vec{OA})\cdot (\vec{PO}-\vec{OA})$

finally expand the dot-product to get

$=\vec{PO}^2-\vec{OA}^2=|\vec{PO}|^2-|\vec{OA}|^2=r^2-r^2=0$