Angle in a Circle: Solving with Linear Algebra

[soopo]

Homework Statement

You have a diameter in a circle, 2R. Show that the angle drawn from the end points of the
diameter to the circle is 90 by Linear Algebra.

The Attempt at a Solution

The situation seems to be E^3.
We get triangle A and B inside the circle such that we have ABCD which has two 90 angles: ABC and ADC.

Let one axis be along AB, one axis along AD and one axis along AC such that
A = (0,0,0), B = (1,0,0), C = (0,0,1) and D = (0,1,0).

We want to show that the dot product between AB and BC is zero
AND the product between AD and DC is zero.

#1:
CB = (-1,0,1)
OB = (-1,0,0)

CB * OB = 1
which is a contradiction to the wanted result.
This would mean that the angle is not 90 which is however false.

 

[LCKurtz]

Homework Statement

You have a diameter in a circle, 2R. Show that the angle drawn from the end points of the
diameter to the circle is 90 by Linear Algebra.

It isn't clear what "by Linear Algebra" means. Can you just look at the plane of the circle, call its center the origin and put xy axes in that plane? Since the radius is 2R, the ends of the diameter are (-R,0) and (0,R). Then take a point (x,y) on the circle and show the appropriate vectors are perpendicular using the dot product.

 

[soopo]

It isn't clear what "by Linear Algebra" means. Can you just look at the plane of the circle, call its center the origin and put xy axes in that plane? Since the radius is 2R, the ends of the diameter are (-R,0) and (0,R). Then take a point (x,y) on the circle and show the appropriate vectors are perpendicular using the dot product.

I get the following solution which gives me a Pythogoran triangle so the two vectors must be at 90 degree angle to each other.

$$\frac {y - 0 } { x + R } = \frac {y} {x + R}$$
and
$$\frac {y - 0 } { x - R } = \frac { y} {x - R }$$

which results in $x^2 + y^2 = R^2$.

So the two vectors are at 90 degree angle at each other at least in E^2.

I am not completely sure how to show the same in E^2.
It should be enough to show the permutation of 3 over 2 for each vectors in the space such that each vector in the space is orthogonal.

 

[Mark44]

So the two vectors are at 90 degree angle at each other at least in E^2.

I am not completely sure how to show the same in E^2.

Did you mean E^3? If so, why do you care? The circle lies in a plane, and the triangle is in the same plane.

It should be enough to show the permutation of 3 over 2 for each vectors in the space such that each vector in the space is orthogonal.

Huh?

 

[Mark44]

I get the following solution which gives me a Pythogoran triangle so the two vectors must be at 90 degree angle to each other.

$$\frac {y - 0 } { x + R } = \frac {y} {x + R}$$
and
$$\frac {y - 0 } { x - R } = \frac { y} {x - R }$$

All that you have done here is to calculate the slopes of the line segments from the endpoints of the diameter to a point (x, y) on the circle. If you are supposed to do this by using "linear algebra," you are probably meant to use vectors, and show that these vectors are perpendicular by showing that their dot product is zero. This is what LCKurtz was saying.

which results in $x^2 + y^2 = R^2$.

So the two vectors are at 90 degree angle at each other at least in E^2.

I am not completely sure how to show the same in E^2.
It should be enough to show the permutation of 3 over 2 for each vectors in the space such that each vector in the space is orthogonal.