Proving that a right angle triangle is always formed when two vectors from either side of a semicircle join at a point on the circles circumferance

Question

The goal is to prove that $\vec{r}_1$ is perpendicular to $\vec{r}_2$.

To begin, I started with the definition that if two vectors are perpendicular to each other, then their scalar product will be zero. For the case of this question, $$ \vec{r}_1 \cdot \vec{r}_2 = 0 $$ Expanding this would give $$ \vec{r}_1 \cdot \vec{r}_2 = r_{1x}r_{2x} + r_{1y}r_{2y} $$ From the diagram, we know that $R + r_x = r_{1x}$, and $R - r_x = r_{2x}$, so \begin{align} \vec{r}_1 \cdot \vec{r}_2 &= (R + r_x)(R - r_x) + r_{1y}r_{2y}\\ &= R^2 - r_x^2 + r_{1y}r_{2y} \end{align} Since $r_y = r_{1y} = r_{2y}$, $$ \vec{r}_1 \cdot \vec{r}_2 = R^2 - r_x^2 + r_y^2 $$ Then substituting in $r_x^2 = r^2 - r_y^2$ \begin{align} \vec{r}_1 \cdot \vec{r}_2 &= R^2 - (r^2 - r_y^2) + r_y^2\\ &= R^2 - r^2 + r_y^2 + r_y^2\\ &= R^2 - r^2 + 2r_y^2 \end{align} Since $\|\vec{r}\| = r = R$ \begin{align} \vec{r}_1 \cdot \vec{r}_2 &= R^2 - R^2 + 2r_y^2\\ &= 2r_y^2 \end{align} But this cannot be so. The scalar product must be equal to $0$. How is it that its equal to $2r_y^2$? Where did I go wrong in my math?

Answer 1

That's $r_{2x} = r_x-R$, not the reverse.

Answer 2

What I am going to show is that if the dot product $\vec{r_1}.\vec{r_2}=0$, then the triangle is a right triangle.

Let us denote by $\vec{R}=\vec{\Omega O}$ where $\Omega$ is the origin of axes and $O$ the center of the circle.

We have:

$$\vec{r_1}+\vec{r_2}=2 \vec{R}$$

Take the scalar product of both sides with themselves:

$$(\vec{r_1}+\vec{r_2})^2=4 \vec{R}^2$$

$$\vec{r_1}^2+\vec{r_2}^2 + 2 \vec{r_1}.\vec{r_2}=4 \vec{R}^2$$

As we have assumed $\vec{r_1}.\vec{r_2}=0$ we see that

$$r_1^2+r_2^2=(2 R)^2$$

which means (reciprocal of Pythagoras' theorem) that we have a right triangle.