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

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

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


    I try with taking $\displaystyle (P_2 - P_1) \cdot (Q - P_0) = 0$ (dot product = 0) but I end up wrong.
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    Re: Line and point in $\mathbb{R}^3$

    What is Q?
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    Re: Line and point in $\mathbb{R}^3$

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

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

    Find $\displaystyle t$ such that $\displaystyle \overrightarrow {{P_0}P} \cdot <2,1,5>=0$
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    Re: Line and point in $\mathbb{R}^3$

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

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

    Find $\displaystyle t$ such that $\displaystyle \overrightarrow {{P_0}P} \cdot <2,1,5>=0$
    I find $\displaystyle \overrightarrow {{P_0}P} = P - P_0 = (1+2t,1+t,1+5t) - (1,0,-2)=(2t,1+t,3+5t).$
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    Re: Line and point in $\mathbb{R}^3$

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

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

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