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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?

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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..................................

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  • $\begingroup$ Ah, I understand the steps here. Just one question, why did you do BA and not AB as we're projecting A onto B? $\endgroup$ – OHAiTHARU Mar 9 '15 at 13:05
  • $\begingroup$ @OHAiTHARU It is just that I prefer to have both vectors involved in the dot product to be "anchored" on the same or on the same line. Since the line's direction vector is "where $\;B\;$ is", I'd rather have it $\;BA\;$ and not the other way around. It doesn't really matter here, of course. $\endgroup$ – Timbuc Mar 9 '15 at 13:07
  • $\begingroup$ @Timbuc: Shouldn't the projection be $\vec B - (1,2,4)$ ? ( The line doesn't pass at $(0,0,0)$, it "starts" at $(1,2,4)$ ) $\endgroup$ – Dor Mar 13 '15 at 14:24
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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: $$ \left(-\dfrac{89}{63}\;;\;2\;;\;\dfrac{44}{63} \right) $$

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