Translation of basis for a vector space on the specified distance

Question

In the Euclidean space $XYZ$ is a basis $X_1Y_1Z_1$ defined that is specified by the vectors $\overrightarrow {O_1X_1}$, $\overrightarrow {O_1Y_1}$ and $\overrightarrow {O_1Z_1}$. How to calculate vectors $\overrightarrow {O_2X_2}$, $\overrightarrow {O_2Y_2}$ and $\overrightarrow {O_2Z_2}$ of the new Basis $X_2Y_2Z_2$ if it is placed on the same line that is defined by the vector $\overrightarrow {O_1Y_1}$ and the distance between the Points $O_1$ and $O_2$ is $d$?

Answer 1

The line $O_1 Y_1$ has parameter equation: $\vec{X}=\vec{O_1}+ t \left( \vec{Y}-\vec{O_1}\right)$. The distance between $O_1$ and $O_2$ is $d$, so $O_2$ the point on the line $O_1 Y_1$ with $$t=\frac{d}{\left| \vec{Y_1}-\vec{O_1} \right|}$$ The absolute value signs stands for the norm of a vector. The translation vector $\vec{O_1 O_2}$ is equal to $$\frac{d}{\left| \vec{Y_1}-\vec{O_1} \right|}(\vec{Y_1}-\vec{O_1})$$ In conclusion: $$\vec{X_2}=\vec{X_1}+\frac{d}{\left| \vec{Y_1}-\vec{O_1} \right|}(\vec{Y_1}-\vec{O_1})$$ $$\vec{Y_2}=\vec{Y_1}+\frac{d}{\left| \vec{Y_1}-\vec{O_1} \right|}(\vec{Y_1}-\vec{O_1})$$ $$\vec{Z_2}=\vec{Z_1}+\frac{d}{\left| \vec{Y_1}-\vec{O_1} \right|}(\vec{Y_1}-\vec{O_1})$$