Question

Given a hexagonal tiling of the plane,such as you might find on a kitchen floor, consider the basisBof${\mathbf{ℝ}}^{\mathbf{2}}$consisting of the vectors$\overrightarrow{\mathbf{v}}\mathbf{,}\overrightarrow{\mathbf{w}}$in the following sketch:

(a) Find the coordinate vectors${\left[\overline{OP}\right]}_{\mathbf{a}}$and${\left[\overrightarrow{OQ}\right]}_{\mathbf{a}}$.Hint: Sketch the coordinate grid defined by the basis$\mathit{B}\mathbf{=}\left(\overrightarrow{V},\overrightarrow{w}\right)$.

(b) We are told that${\left[\overrightarrow{\mathrm{OR}}\right]}_{\mathbf{B}}\mathbf{=}\left[\begin{array}{c}3\\ 2\end{array}\right]$. Sketch the pointR. IsRa vector or a center of a tile?

(c) We are told that${\left[\overrightarrow{\mathrm{OS}}\right]}_{\mathbf{B}}\mathbf{=}\left[\begin{array}{c}17\\ 13\end{array}\right]$. IsSa center or a vertex of a tile?

Short answer

Thus, (a) the value of${\left[\overrightarrow{OP}\right]}_B=\overrightarrow{v}+\overrightarrow{w}$and${\left[\overrightarrow{OQ}\right]}_B=2\overrightarrow{v}+\overrightarrow{w}$.

(b) The sketch is obtained above representingRas a centre.

(c) it is proved that Sis a vertex of tile.

Step-by-step solution

(a) Determine the values of [OP→]B and [OQ→]B.

For a subspace with basis$B=\overrightarrow{v_1},\overrightarrow{v_2},\dots ,\overrightarrow{v_n}$the vector will be$\overrightarrow{\mathbf{x}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\overrightarrow{{\mathbf{v}}_{\mathbf{1}}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\overrightarrow{{\mathbf{v}}_{\mathbf{2}}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{c}}_{\mathbf{n}}\overrightarrow{{\mathbf{v}}_{\mathbf{n}}}\mathbf{.}$.

Now, let write as follows:

$Vertex\hspace{0.33em}{P}^{\text{'}}=vector\hspace{0.33em}\overrightarrow{v}$connects with O.

$Vertex\hspace{0.33em}{Q}^{\text{'}}=vector\hspace{0.33em}\overrightarrow{w}$connects with O.

As ${\left[P^{i}P\right]}_{\mathcal{B}}$and ${\left[P^{i}P\right]}_g$are parallel so, write as follows:

$$\begin{gathered} {\left[P^{i}P\right]}_{\mathcal{B}}=\overrightarrow{v}+\overrightarrow{w} \\ ={\left[Q^{\overrightarrow{T}Q}\right]}_{؏} \end{gathered}$$

If take$2\overrightarrow{v}+2\overrightarrow{w}$as get a diagonal of a regular hexagon which is two times larger then sides parallel with it.

$$\begin{gathered} {\left[\overrightarrow{OP}\right]}_B={\left[P^{i}P\right]}_{\mathfrak{R}} \\ =\overrightarrow{v}+\overrightarrow{w} \end{gathered}$$

And

$$\begin{gathered} {\left[\overrightarrow{OQ}\right]}_B=\overrightarrow{v}+{\left[P^{i}P\right]}_{\mathfrak{R}} \\ =\overrightarrow{v}+\overrightarrow{v}+\overrightarrow{w} \\ =2\overrightarrow{v}+\overrightarrow{w} \end{gathered}$$

Hence, the value of ${\left[\overrightarrow{OP}\right]}_B=\overrightarrow{v}+\overrightarrow{w}$and ${\left[\overrightarrow{OQ}\right]}_B=2\overrightarrow{v}+\overrightarrow{w}$.

(b) Determine the Sketch of the point .

Let assume that hexagonal tile of the plane. Consider the basis B of$R^2$consist of the vector$\overrightarrow{v},\overrightarrow{w}$.

Also, it is given that${\left[\overrightarrow{OR}\right]}_B=\left[\begin{array}{c}3\\ 2\end{array}\right]$.

$Vertex\hspace{0.33em}{P}^{\text{'}}=vector\hspace{0.33em}\overrightarrow{v}$connects with O.

$Vertex\hspace{0.33em}{Q}^{\text{'}}=vector\hspace{0.33em}\overrightarrow{w}$connects with O.

As${\left[P^{\pm }P\right]}_{kg}$and${\left[P^{i}P\right]}_{iB}$are parallel.

Thus, write as follows:

$$\begin{gathered} {\left[P^{\overline{z}}P\right]}_x=\overrightarrow{v}+\overrightarrow{w} \\ ={\left[QQ\right]}_z \end{gathered}$$

According to the question sketch as follows:

Hence, the sketch is given above representing R as a centre.

(c) Show that S is vertex.

Let assume that hexagonal tile of the plane. Consider the basis B of$R^2$consist of the vector$\overrightarrow{v},\overrightarrow{w}$.

Also, it is given that${\left[\overrightarrow{OS}\right]}_B=\left[\begin{array}{c}17\\ 13\end{array}\right]$.

According to the question,

$$\begin{gathered} \overrightarrow{v}+\overrightarrow{w}=Center \\ 2\left(\overrightarrow{v}+\overrightarrow{w}\right)=Vertex \\ 3\left(\overrightarrow{v}+\overrightarrow{w}\right)=Vertex \end{gathered}$$

Thus, it gives as follows:

$$\begin{gathered} 3k\left(\overrightarrow{v}+\overrightarrow{w}\right)=Vertex \\ 3k\left(\overrightarrow{v}+\overrightarrow{w}\right)+\left(\overrightarrow{v}+\overrightarrow{w}\right)=Center \end{gathered}$$

By interpretation it gives;

$$\begin{gathered} 13\left(\overrightarrow{v}+\overrightarrow{w}\right)=3.4\left(\overrightarrow{v}+\overrightarrow{w}\right)+\left(\overrightarrow{v}+\overrightarrow{w}\right) \\ =Center \end{gathered}$$

Now, solve as follows:

$$\begin{gathered} 17\overrightarrow{v}+13\overrightarrow{w}=4\overrightarrow{v}+13\left(\overrightarrow{v}+\overrightarrow{w}\right) \\ =4\overrightarrow{v}+Center \\ =Vertex \end{gathered}$$

According to the given question, the sketch is given below:

Hence, S is a vertex of tile.