A question Of 3D Geometry.

Question

The Co-ordinates of the foot of the perpendicular drawn from the point $P(1,0,3)$ to the join of the points $A(4,7,1)$ and $B(3,5,3)$ is?

I need help finding the equation of $AB$ in the form $ax+by+cz+d=0$ and then i think i can proceed from there on my own. Is there any other method by which we can find the answer to it?

Answer 1

$\text{The way I did it:}$
$\text{All three of the given points: A, B, C}$
$\text{reside on the plane }$
$10x-4y+z=13.$
$\text{The altitude from P cuts the extension}$
$\text{of the line AB at W.}$
$\text{The analog of ABP is abp.}$

$\begin{array}{l|l} A(4\,|\,7\,|\,1)&a(0\,|\,0\,|\,0)\\ B(3\,|\,5\,|\,3)&b(3\,|\,0\,|\,0)\\ P(1\,|\,0\,|\,3)&p(7\,|\,\sqrt{13}\,|\,0)\\ W(\frac53\,|\,\frac73\,|\,\frac{17}3)&w(7\,|\,0\,|\,0) \end{array}$

$\text{AB and ab are 3 units long.}$
$\text{BP and bp are }\sqrt{29}\text{ units long.}$
$\text{PA and pa are }\sqrt{62}\text{ units long.}$

$\text{Obviously, }\overline{aw}=\frac73·\overline{ab},$
$\text {so }\overline{AW}=\frac73·\overline{AB}.$
$\text{W is easily located at }$
$(\frac53\mid\frac73\mid\frac{17}3).$

Answer 2

HINT: let $P'$ the searched footpoint of $P$ then $P'$ must situated on $AB$ and the dot product $$\vec{AB}\cdot \vec{P'P}=0$$ can you get it from here? the straight line $AB$ has the equation $$[x,y,z]=[4;7;1]+t[-1;-2;2]$$ and $t$ is a real number $$\vec{P'P}=[1-x;-y;3-z]$$ and $$[1-x;-y;3-z]\cdot [-1;-2;2]=0$$ and $$x=4-t;y=7-2t;z=1+2t$$