HELP! Find the point on the line closest to Q.

[maeveoneill]

The question is:
Find the point R on l with parametric equations : x= 1-2t, y= 1, z= 1+3t that is closets to the point Q= (0,1,0)

I am using projections to solve this.
I found that v = PQ= [1 0 1]
projd(v)= [-2/13 0 3/13]
v- projd(v)= [15/3 0 10/3]

where v-projd(v) is the vector from P to Q.

From here I am stuck. I was thinking I could use the parametric equations of the line.. and the parametric equations of the vector v-projd(v). The intersection of both of these would give me the point R.
Is this right? and how do I find the point of intersection using parametric equations or the vector form of a line?

Please help!

 

[soroban]

Hello, maeveoneill!

\(\displaystyle \text{Find the point }R\text{ on line }L\text{ with parametric equations: }\:\begin{Bmatrix}x&=& 1-2t \\ y&=& 1 \\ z&=& 1+3t\end{Bmatrix}\)
. . \(\displaystyle \text{that is closest to the point }Q\,(0,1,0)\)

\(\displaystyle \text{Line }L\text{ has direction vector: }\:\vec u \:=\:\langle -2,\:0,\:3\rangle\)


\(\displaystyle \text{Let }R\text{ be any point on line }L\!:\;\;R(1-2t,\:1,\:1+3t)\)

\(\displaystyle \text{The vector from }Q\text{ to }R\text{ is: }\:\vec v \:=\:\overrightarrow{QR}\:=\:\langle 1-2t,\:0,\:1+3t\rangle\)


\(\displaystyle \text{We want: }\:\vec u \perp \vec v\)

. . \(\displaystyle \text{That is: }\:\vec u\cdot\vec v \:=\:0 \quad\Rightarrow\quad \langle -2,\,0,\,3\rangle\cdot\langle 1-2t,\,0,\,1+3t\rangle \:=\:0\)

. . \(\displaystyle \text{We have: }\:-2(1-2t) + 0 + 3(1+3t) \:=\:0 \quad\Rightarrow\quad13t+1\:=\:0 \quad\Rightarrow\quad t \,=\,-\tfrac{1}{13}\)


\(\displaystyle \text{Therefore: point }R\text{ is: }\:\bigg(1-2\left[\text{-}\tfrac{1}{13}\right],\:1,\:1+3\left[\text{-}\tfrac{1}{13}\right]\bigg) \;=\; \bigg(\tfrac{15}{13},\:1,\:\tfrac{10}{13}\bigg)\)

 

[maeveoneill]

That makes sense, and seems even easier.. but I think this is to be solved using projections as I had started. is that possible??

 

[Deleted member 4993]

That makes sense, and seems even easier.. but I think this is to be solved using projections

You are using scalar (dot) product - and projections are based on that.

as I had started. is that possible??