Question

1. let (a, b \in \mathbb{r}^2) and let (\theta = 180^\circ). prove that (r_{b,\theta}r_{a,\theta}) is a translation through vector (2\overrightarrow{ab}).

Explanation:

⚡ Using what you learned: Compositions of Transformations

Step 1: Define the rotation formula

A rotation of \(180^\circ\) about a point \(C\) is a half-turn. For any point \(P \in \mathbb{R}^2\), its image \(P' = R_{C, 180^\circ}(P)\) satisfies:

$$ \frac{P + P'}{2} = C \implies P' = 2C - P $$

Step 2: Apply the first rotation

Let \(P\) be an arbitrary point in \(\mathbb{R}^2\). Apply the rotation \(R_{A, \theta}\) about point \(A\) with \(\theta = 180^\circ\):

$$ P' = R_{A, 180^\circ}(P) = 2A - P $$

Step 3: Apply the second rotation

Apply the rotation \(R_{B, \theta}\) about point \(B\) with \(\theta = 180^\circ\) to the image \(P'\):

$$ P'' = R_{B, 180^\circ}(P') = 2B - P' $$

Substitute \(P' = 2A - P\) into the equation:

$$ P'' = 2B - (2A - P) = P + 2(B - A) $$

Step 4: Express as a translation

The vector from point \(A\) to point \(B\) is defined as:

$$ \overrightarrow{AB} = B - A $$

Substituting this into the expression for \(P''\):

$$ P'' = P + 2\overrightarrow{AB} $$

This represents a translation of the point \(P\) by the vector \(2\overrightarrow{AB}\). Thus, \(R_{B, \theta}R_{A, \theta}\) is a translation through vector \(2\overrightarrow{AB}\).

Answer:

By representing the \(180^\circ\) rotations about points \(A\) and \(B\) algebraically:

1. \(R_{A, 180^\circ}(P) = 2A - P\)
2. \(R_{B, 180^\circ}(R_{A, 180^\circ}(P)) = 2B - (2A - P) = P + 2(B - A)\)

Since \(B - A = \overrightarrow{AB}\), the composition simplifies to:

$$ (R_{B, 180^\circ} \circ R_{A, 180^\circ})(P) = P + 2\overrightarrow{AB} $$

This proves that the composition of the two rotations is a translation by the vector \(2\overrightarrow{AB}\).