Problem on geometric proofs with vectors
$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
Show that the lines connecting any point on the semicircle of radius 1 to $(1,0)$ and $(-1,0)$ are perpendicular.
Solution
Our strategy will be to express the two line segments as vectors and to show that these vectors are perpendicular.
We let $\bfx$ be any point on the semicircle. The two ends of the semicircle can be expressed in vector form as $\bfi$ and $- \bfi$:
Now we express the vectors from $\pm \bfi$ to $\bfx$. Recall that:
Subtraction gives the vector between two points
The vector from $\mathbf{y}$ to $\mathbf{x}$ is given by $\mathbf{x} - \mathbf{y}$.
Hence the two line segments are given by the vectors $\mathbf{x} - (-\mathbf{i}) = \mathbf{x} + \mathbf{i}$ and $\mathbf{x} - \mathbf{i}$:
As we would like to show that these vectors are perpendicular, we recall
Dot product of perpendicular vectors is zero
Two vectors are perpendicular if their dot product is zero.
Algebra of dot products
Hence, we compute
\begin{align}
\left( \mathbf{x} - \mathbf{i} \right) \cdot \left( \mathbf{x} + \mathbf{i} \right) &= \mathbf{x} \cdot \mathbf{x} + \mathbf{x} \cdot \mathbf{i} - \mathbf{i} \cdot \mathbf{x} - \bfi \cdot \bfi \\
&= \bfx \cdot \bfx - 1
\end{align}
Recalling that
Dot product and vector length
The dot product of a vector with itself equals the squared length of the vector: $$\bfx \cdot \bfx = \left | \bfx \right |^2$$
Hence, we have $$\left( \mathbf{x} - \mathbf{i} \right) \cdot \left( \mathbf{x} + \mathbf{i} \right) = \left | \bfx \right |^2 - 1.$$
Because $\mathbf{x}$ is on the unit semicircle, $\left| \mathbf{x} \right| = 1$, hence $\left( \mathbf{x} - \mathbf{i} \right) \cdot \left( \mathbf{x} + \mathbf{i} \right) = 0$, and the line segments are perpendicular.
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