Question

a.Suppose that a and b are position vectors. Prove that (r-a).(r-b)=0 is the equation of a sphere, where r is the position vector of any point on the sphere.
b. Where are a and b located relative to the sphere?

Answer 1

@wio

Answer 2

What do we have to do here?
Should we have to find Where are a and b located relative to the sphere?

Answer 3

Remember that \(\mathbf{r}=<x,y>\)

Answer 4

@wio here
\[r = <x , y , z >\]
as the problem mentioned sphere , not a circle!

Answer 5

Let \(\mathbf{r}=<x,y,z>,\;\mathbf{a}=<a_x,a_y,a_z>,\;\mathbf{b}=<b_x,b_y,b_z>\)
Just manually do the vector subtraction and dot product.

Answer 6

okay

Answer 7

I try to expand it

Answer 8

and then ? :$

Answer 9

Umm, use something like difference of squares to make it look like an equation of a sphere.

Answer 10

What does the equation currently look like?

Answer 11

Shall i tell the equation?

Answer 12

@SandeepReddy I'm asking @pinky_cute1995 where she is currently at.

Answer 13

Okay , Good Luck @pinky_cute1995 !

Answer 14

ohh, it's weird lmao

Answer 15

I can't cancel anything out

Answer 16

I did what u said (:

Answer 17

Can you type your current equation? Nothing is supposed to cancel.

Answer 18

ohh really, I thought we have to prove if left and right side are the same ?

Answer 19

No, we have to prove it is a sphere.
We need a equation like: \[
(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2
\]

Answer 20

what she should have now is
\[\left(\begin{matrix}\ x^2-xa_{x}-xb_{x}+a_{x}b_{x}\ \end{matrix}\right)\]
and the same thing with y and z

Answer 21

all equals 0

Answer 22

is this one is the equation of a sphere: xdx +ydy +zdz =0

Answer 23

wio gave the equation for a sphere above.

Answer 24

ok, maybe we're approaching it the wrong way.
Let's try the other way. we can see that r-a is perpendicular to r-b
@pinky_cute1995 you see why?

Answer 25

yeah

Answer 26

Here is another macro!

Answer 27

let's think about a simpler case. imagine just one circle in the sphere

Answer 28

a, b, and r form a triangle in this circle.

Answer 29

okayy

Answer 30

I remember something like two vector in a circle create a 90 degree angle ?

Answer 31

think back to geometry when you were working with circles and triangles.

Answer 32

Sorry, experimenting on this extension and it's causing weird behavior...

Answer 33

hint:

Answer 34

it's one specific insight you have to see here. if you're not getting it I could just tell you and all you would lose is the "AHA!" moment.

Answer 35

?

Answer 36

I gonna go to sleep. :O

Answer 37

ok. A and B have to be on the ends of a diameter of the circle.

Answer 38

okay

Answer 39

for any point r, the angle ra and rb will be a right angle.

Answer 40

if you extend this to 3 dimensions, you get a sphere

Answer 41

just prove it like that ?

Answer 42

draw the circle and the triangle ?

Answer 43

you should probably say something about thales' theorem, but yes.

Answer 44

but what are vector a, b and r ?

Answer 45

cos they give us vector (r-a).(r-b) isn't it

Answer 46

a and b are points on the ends of a diameter of the sphere
r is any other point on the sphere

Answer 47

the only things (r-a).(r-b)=0 can give you are a headache if you try and work it out numerically and the fact that segments ra and rb are perpendicular

Answer 48

cos they said they are vectors

Answer 49

position vectors, which means that they represent points. The position vector is the distance as you move from the origin to the point.

Answer 50

ohh, okay I see

Answer 51

I can't think at midnight

Answer 52

I understand that.

Answer 53

thank you anyway

Answer 54

you are lucky I got on now, i'm not usually here at this time of day.

Answer 55

goodnight

Answer 56

hahaha, I am (: