The crystal lattice for titanium has the hexagonal structure shown in the accompanying figure. The vectors $\mathbf{b}_{1}=\left[\begin{array}{r}2.6 \ -1.5 \ 0.0\end{array}\right] \cdot \mathbf{b}_{2}=\left[\begin{array}{r}0.0 \ 3.0 \ 0.0\end{array}\right] \cdot \mathbf{b}_{3}=\left[\begin{array}{r}0.0 \ 0.0 \ 4.8\end{array}\right]$ in $\mathbf{R}^{3}$ form a basis for the unit cell shown in the accompanying figure. One of the octahedral sites is $\left[\begin{array}{lll}-\frac{1}{2} & \frac{1}{4} & \frac{1}{6}\end{array}\right]^{\top}$, relative to the lattice basis. Determine the coordinates of this site relative to the standard basis of $\mathrm{R}^{3}$.
Click the icon to view the figure.

Determine the coordinates of this site relative to the standard basis of $\mathbf{R}^{3}$.
$$
\mathbf{x}=\square
$$
(Type an integer or decimal for each matrix element.)

Question

The crystal lattice for titanium has the hexagonal structure shown in the accompanying figure. The vectors $\mathbf{b}_{1}=\left[\begin{array}{r}2.6 \ -1.5 \ 0.0\end{array}\right] \cdot \mathbf{b}_{2}=\left[\begin{array}{r}0.0 \ 3.0 \ 0.0\end{array}\right] \cdot \mathbf{b}_{3}=\left[\begin{array}{r}0.0 \ 0.0 \ 4.8\end{array}\right]$ in $\mathbf{R}^{3}$ form a basis for the unit cell shown in the accompanying figure. One of the octahedral sites is $\left[\begin{array}{lll}-\frac{1}{2} & \frac{1}{4} & \frac{1}{6}\end{array}\right]^{\top}$, relative to the lattice basis. Determine the coordinates of this site relative to the standard basis of $\mathrm{R}^{3}$.
Click the icon to view the figure.

Determine the coordinates of this site relative to the standard basis of $\mathbf{R}^{3}$.
$$
\mathbf{x}=\square
$$
(Type an integer or decimal for each matrix element.)

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the coordinates of the octahedral site relative to the standard basis, we need to multiply the given basis vectors by the site's coordinates in the lattice basis. ‖ ‖ step 2 ⋮ The multiplication of the matrix formed by the basis vectors $\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3$ and the column vector of the site's coordinates is calculated as follows: $(2.6 | -1.5 | 0.0 \quad 0 | 3 | 0 \quad 0 | 0 | 4.8) \cdot \left\{-\frac{1}{2}, \frac{1}{4}, \frac{1}{6}\right\}$. ‖ ‖ step 3 ⋮ The result of the multiplication gives us the coordinates of the site relative to the standard basis: $\{-1.675, 0.75, 0.8\}$. ‖ ‖ step 4 ⋮ Summing up the individual coordinates gives us the final coordinate of the site relative to the standard basis: $-1.675 + 0.75 + 0.8 = -0.125$. However, this step seems to be an error because we should not sum the coordinates. ‖ ‖ step 5 ⋮ The correct coordinates of the site relative to the standard basis are given by the vector: $(-0.836651, 0.37462, 0.399594)$. ‖ ∻Answer∻ ⚹ (-0.836651, 0.37462, 0.399594) ⚹ ∻Key Concept∻ ⚹Matrix-vector multiplication to convert coordinates from one basis to another⚹ ∻Explanation∻ ⚹The coordinates of a point relative to a new basis are found by multiplying the matrix of the new basis vectors by the point's coordinate vector relative to that basis. The result is the coordinates of the point in the standard basis.⚹◊What is the formula to convert coordinates from a given basis to the standard basis in $ \mathbf{R}^{3} $?⍭ Generate me a similar question◊

Sia