# You MUST use vectors operations and/or dot product to solve this question. Any other method will not be accepted. Let A and B be the endpoints of a diameter of a circle. Let C be any point of the...

You **MUST** use vectors operations and/or dot product to solve this question. Any other method will not be accepted.

Let A and B be the endpoints of a diameter of a circle. Let C be any point of the same circle. Show that the segments CA and CB are perpendicular.

Hint: I recommend using vector v =vector AM and vector w = vector MC

### 1 Answer | Add Yours

The middle is M; the tip of the red arrow is A; the tip of the blue arrow is B; the tip of all the other arrows is C

So:

red arrow: `vec(MA)=-vec(v)`

blue arrow: `vec(MB)=vec(v)`

green arrow: `vec(MC)=vec(w)`

purple arrow: `vec(BC)`

orange arrow: `vec(AC)`

notice:

green = red+orange, so:

orange = green - red

purple = (-blue) + green

so:

orange = green-red = `vec(w)-(-vec(v))`

purple = (-blue)+green = `-vec(v)+vec(w)`

To show that the orange and purple vectors are perpendicular, we take the dot product and show it is zero:

`(vec(w)+vec(v)) * (vec(w)-vec(v))=`

`||vec(w)||^2 - vec(v)*vec(w) + vec(v)*vec(w) - ||vec(v)||^2 =`

`||vec(w)||^2 - ||vec(v)||^2 `

but these are the same length because both `vec(v)` and `vec(w)` are radii, and so have the same length

`||vec(w)||^2 - ||vec(v)||^2 =0`

So the vectors are perpendicular