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

• Nov 22nd 2013, 12:02 AM
jacob93
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.
• Nov 22nd 2013, 12:25 AM
Prove It
Re: Line and point in $\mathbb{R}^3$
What is Q?
• Nov 22nd 2013, 06:10 AM
Plato
Re: Line and point in $\mathbb{R}^3$
Quote:

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 $\overrightarrow {{P_0}P} \cdot <2,1,5>=0$
• Nov 24th 2013, 05:44 AM
jacob93
Re: Line and point in $\mathbb{R}^3$
Quote:

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 $\overrightarrow {{P_0}P} = P - P_0 = (1+2t,1+t,1+5t) - (1,0,-2)=(2t,1+t,3+5t).$
• Nov 26th 2013, 03:36 AM
jacob93
Re: Line and point in $\mathbb{R}^3$
Quote:

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 $t=-\frac{16}{30}$