Linear Algebra: Distance From a Point and a Line in $R^3$

Question

Question: Find the distance from the point $B = (1, 1, 1)$ and the line l through the point $A = (-1, 1, 2)$ and the direction vector $d = [1, 0, 1]$.

Here is what I've found: Vector Form of line l (maybe?): $[-1, 1, 2] + t[1, 0, 1] = [-1+t, 1, 2+t] =$ I call it the vector $x$ And the parametric equations: $x = -1 + t$

$y = 1$

$z = 2 + t$

I think what I need to do is find the projection of vector b onto vector x which would be:

$(x•b)/(x•x) * x$ ...or its also equal to: $(x•b)/(|x|^2) * x$

So if all of my thinking is correct, my only problem is I don't know how to do a projection in $R^3$. Especially not with a vector in the parametric form which is all I can seem to get it in. Is there any advice, or a reference you can point me towards - I haven't been able to find any so far ...

Answer 1

$D(t)=\sqrt{({2-t})^2+0^2+({-1-t})^2}=\sqrt{4-4t+t^2+1+2t+t^2}=\sqrt{2t^2-2t+5}$.

To simplify matters, let's minimize $D^2=2t^2-2t+5$...

$D'(t)=4t-2=0\implies t=\frac12$.

Now evaluate $D$ at $\frac12$: $$D(\frac 12)=\sqrt{2\frac14-1+5}=\sqrt{\frac92}=\pm\frac3{\sqrt2}$$.

Answer 2

How about minimizing the distance from $l$ to $B$? If you know any calculus, you know this can be done by setting the derivative of the distance function (a function of $t$), equal to zero...

As for projection, you seem to have the right formula. When you cover Graham-Schmidt you will get good at this...

So, $(B-A)\cdot d=(2,0,-1)\cdot(1,0,1)=1$. Divide by $\lvert d\rvert =\sqrt2$. Now by the Pythagorean theorem, we have ${\lvert B-A\rvert} ^2=(\frac1{\sqrt2})^2+D^2$. So $5=\frac12+D^2$. So $D=\frac3{\sqrt2}$...