## Materials Science and Engineering: An Introduction

The point is $(4,3,6)$
Given that the direction indices for a vector that passes from point it represent the same rhombohedral unit cell is same as tetragonal unit cell $\frac{1}{3},\frac{1}{2},0$ to the point $\frac{2}{3},\frac{3}{4},\frac{1}{2}$ So the point is $x_{1}=\frac{a}{3}$ $y_{1}=\frac{b}{2}$ $z_{1}=0c=0$ Similarly the point 2 is $x_{2}=\frac{2a}{3}$ $y_{2}=\frac{3b}{4}$ $z_{2}=\frac{c}{2}$ Equation representation of u,v and w $u=n(\frac{x_{2}-x_{1}}{a})$ $v=n(\frac{y_{2}-y_{1}}{b})$ $z=n(\frac{z_{2}-z_{1}}{c})$ Denominator in the fraction has highest value of 3 and 4. So the value of n =12 $u=12(\frac{\frac{2a}{3}-\frac{a}{3}}{a})=4$ $v=12(\frac{\frac{3b}{4}-\frac{b}{2}}{b})=3$ $z=12(\frac{\frac{c}{2}-0}{c})=6$ The point is $(4,3,6)$ .