Projection of a point onto a line I am given the line $l:\langle x,y,z\rangle = \langle1,2,4\rangle + \lambda\langle-1,0,8\rangle, \lambda\in \mathbb{R}$.
I am asked to find the projection (as a vector) of the point, $A\equiv(1,1,1)$ onto the line $l$.
I know the equation for projection of a vector onto another vector, and I can find the projection of a point onto a line that would give me a point as the end result. I do not know how to find the projection of a point onto a line (in that form) that would give me a vector as a result. Is there a formula/series of steps to do this?
 A: We need a point $\;B\;$ on the line s.t. $\;\vec{BA}\perp(-1,0,8)\;$ (why?) . Since any such point $\;B\;$ is of the form $\;(-t+1\,,\,2\,,\,8t+4)\;,\;\;t\in\Bbb R\;$ , we need to solve the equation
$$0=\vec{BA}\cdot(-1,0,8)=(t,-1,-8t-3)\cdot(-1,0,8)=-t-64t-24$$
and then substitute the obtained value of $\;t\;$ back in the expression for $\;B\;$
Cofusing stuff here..................................
A: The points of the line have coordinates:
$$
(x,y,z)=(1-\lambda\;;\;2\;;\;4(1+2\lambda))=(x\;;\;2\;;\;4(3+2x))
$$
The plane perpendicular to the line and passing thorough $A$ has equation:
$$
\langle (-1;0;8),(x-1;y-1;z-1) \rangle=0 \Rightarrow -x+8z-7=0
$$
The orthogonal projection of $A$ on the stright line is the intersection of this plane with the line. Noting that the points of the plane are:
$$
\left(x;y\;;\; \dfrac{7+x}{8}\right)
$$
we have:
$$
\dfrac{7+x}{8}=4(3+2x) \Rightarrow x= -\dfrac{89}{63}
$$
so the searched point is:
$$
\left(-\dfrac{89}{63}\;;\;2\;;\;\dfrac{44}{63} \right)
$$ 
