# Thread: Line and point in $\mathbb{R}^3$

1. ## Line and point in $\mathbb{R}^3$

Let $P_0 = (1,0,-2)$ and let $\ell$ be the line that goes through $P_1 = (1,1,1)$ and $P_2 = (3,2,6)$. Find the coordinates for a point $P$ on $\ell$ such that $P_0$ and $P$ are orthogonal to $\ell$.

I try with taking $(P_2 - P_1) \cdot (Q - P_0) = 0$ (dot product = 0) but I end up wrong.

What is Q?

3. ## Re: Line and point in $\mathbb{R}^3$

Originally Posted by jacob93
Let $P_0 = (1,0,-2)$ and let $\ell$ be the line that goes through $P_1 = (1,1,1)$ and $P_2 = (3,2,6)$. Find the coordinates for a point $P$ on $\ell$ such that $P_0$ and $P$ are orthogonal to $\ell$.
The line $\ell(t)=(1+2t,1+t,1+5t)$ so $P\in\ell$ means $P=(1+2t,1+t,1+5t)$ for some $t$.

So $\overrightarrow {{P_0}P} = \left( {2t,t, - 1 + 5t} \right)$.

Find $t$ such that $\overrightarrow {{P_0}P} \cdot <2,1,5>=0$

4. ## Re: Line and point in $\mathbb{R}^3$

Originally Posted by Plato
The line $\ell(t)=(1+2t,1+t,1+5t)$ so $P\in\ell$ means $P=(1+2t,1+t,1+5t)$ for some $t$.

So $\overrightarrow {{P_0}P} = \left( {2t,t, - 1 + 5t} \right)$.

Find $t$ such that $\overrightarrow {{P_0}P} \cdot <2,1,5>=0$
I find $\overrightarrow {{P_0}P} = P - P_0 = (1+2t,1+t,1+5t) - (1,0,-2)=(2t,1+t,3+5t).$

5. ## Re: Line and point in $\mathbb{R}^3$

Originally Posted by Plato
The line $\ell(t)=(1+2t,1+t,1+5t)$ so $P\in\ell$ means $P=(1+2t,1+t,1+5t)$ for some $t$.

So $\overrightarrow {{P_0}P} = \left( {2t,t, - 1 + 5t} \right)$.

Find $t$ such that $\overrightarrow {{P_0}P} \cdot <2,1,5>=0$
I get $t=-\frac{16}{30}$