What is the shortest distance between the point $(1,8,-4)$ and the line that is given by the equations,
$-2/5+(1/5)x=-2+(1/2)y=-3-(1/2)z$?
My guess it that it has to do with a straight line.. Because the shortest distance between tho points A and B is a straight line. My idea was to implement the dot product or the projection formula in some way. I could find a point on the line by plugging in for example, $x=2$, obtaining an answer and solving for $y$ and $z$. This gets me a point on the line $(2,4,-6)$. I could find the distance between the point on the line and the point outside the line with the assistance of the distance formula, which assetains me, $\sqrt{21}$. Now, I would wish to be able to project $\sqrt{21}$ on the line but I struggle. This is where I need help, thanks in advance.
If you can write te equation of the line in vector form, there is already a formula how to calculate this distance. See for example here. So we write $\vec r=\vec r_0+\vec n t$. If we write this by components we have $$\begin{align}x&=x_0+n_x t\\y&=y_0+n_y t\\z&=z_0+n_z t\end{align}$$ You can now explicitly calculate $t$ from every equation, and you get $$\frac{x-x_0}{n_x}=\frac{y-y_0}{n_y}=\frac{z-z_0}{n_z}$$ It should be now trivial to calculate $\vec r_0$ and $\vec n$ by just identifying the coefficients from your equation with the above formula. Now use the expression on wikipedia to get the perpendicular vector, and then you get the distance by taking the norm.