Generalization of straight line motion under constant acceleration My question is that, we all know the three equations of straight line motion under constant acceleration,
\begin{align}
x & =x_{\rm o}+v_{\rm o}\,t+\tfrac12 \mathrm a\,t^2
\tag{1d-a}\label{1d-a}\\
v & =v_{\rm o}+\mathrm a\,t
\tag{1d-b}\label{1d-b}\\
v^2 & =v_{\rm o}^2+2\,\mathrm a\left(x-x_{\rm o}\right)
\tag{1d-c}\label{1d-c}
\end{align}
Is my generalization correct?
\begin{align}
\mathbf r & =\mathbf r_{\rm o}+\boldsymbol v_{\rm o}\,t+\tfrac12\mathbf a\,t^2 \quad \text{(no difference with that)}
\tag{3d-a}\label{3d-a}\\
\boldsymbol v & =\boldsymbol v_{\rm o}+\mathbf a\,t
\tag{3d-b}\label{3d-b}\\
\vert\boldsymbol v\vert^2 & =\vert\boldsymbol v_{\rm o}\vert^2+2\,\mathbf a\boldsymbol \cdot\left(\mathbf r-\mathbf r_{\rm o}\right)
\tag{3d-c}\label{3d-c}
\end{align}
Please explain the general principle of generalization of 1-dimension formulas into 3-dimensions.
And I must add I am very sorry that I print this question not using LaTeX, I really know nothing about it, so I printed it like that, hopefully you will be patient about me.
 A: Though the above generalisations are correct, what we do in most cases is that we resolute (or break) any given motion along mutually perpendicular axes (namely x, y and z axes) and then apply these formula separately along each of the axes as: $$v_x=v_{0x}+a_{0x}t$$ $$x=v_{0x}t+\frac{1}{2}a_{0x}t^2$$ $$v_x^2=v_{0x}^2+2a_{0x}x$$ along x-axes. Similarly, $$v_y=v_{0y}+a_{0y}t$$ $$y=v_{0y}t+\frac{1}{2}a_{0y}t^2$$ $$v_y^2=v_{0y}^2+2a_{0y}y$$ and $$v_z=v_{0z}+a_{0z}t$$ $$z=v_{0z}t+\frac{1}{2}a_{0z}t^2$$ $$v_z^2=v_{0z}^2+2a_{0z}z$$ along y and z axes respectively. The reason why we do is because motion along mutually perpendicular axes are independent of each other and hence we can apply these formula separately along the axes.
Hope it helps.
A: 
As shown in Figure-01
\begin{align}
\mathbf r & =\mathbf r_{\rm o}+\boldsymbol v_{\rm o}\,t+\tfrac12\mathbf a\,t^2
\tag{01a}\label{01a}\\
\boldsymbol v & =\boldsymbol v_{\rm o}+\mathbf a\,t
\tag{01b}\label{01b}\\
\vert\boldsymbol v\vert^2 & =\vert\boldsymbol v_{\rm o}\vert^2+2\mathbf a\boldsymbol \cdot\left(\mathbf r-\mathbf r_{\rm o}\right)
\tag{01c}\label{01c}
\end{align}
