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Math Help - three questions. 3 plane

  1. #1
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    three questions. 3 plane

    3.a line L in 3 space has vector equation X=(3,2,1)+ a(1,2,2). Find the distance
    from P=(1,0,1) to line L
    I tried use this this formula to solve d=(ax+by+cz-d)/sqrt root (a^2+b^2+c^2)
    but its wrong

    2,Two planes are perpendicular if they have perpendicular normal vectors. find a cartesian equations for a plane which passes through point P=(1,2,3) and is perpendicular to plane 2x+2y-z=1 and also to plane x+y-1=3.


    3,The point A=(1,2,1,1),B=(2,3,0,3),C=(1,1,2,-1) in R4 from a triangle in R4. find a point D in R4 such ABCD is a parallelogram.


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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by questionboy View Post
    I tried use this this formula to solve d=(ax+by+cz-d)/sqrt root (a^2+b^2+c^2)
    but its wrong
    That is a wrong formula. Use the following, if

    x,y,z)=(x_1,y_1,z_1)+t(a,b,c)" alt="P=(x_0,y_0,z_0)\;,\quad Lx,y,z)=(x_1,y_1,z_1)+t(a,b,c)" />

    then,

    d(P,L)=\dfrac{\textrm{mod}\begin{vmatrix}{\vec{i}}  &{\vec{j}}&{\vec{k}}\\{x_1-x_0}&{y_1-y_0}&{z_1-z_0}\\{a}&{b}&{c}\end{vmatrix}}{\sqrt{a^2+b^2+c^2}  }


    Fernando Revilla
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by questionboy View Post
    2,Two planes are perpendicular if they have perpendicular normal vectors. find a cartesian equations for a plane which passes through point P=(1,2,3) and is perpendicular to plane 2x+2y-z=1 and also to plane x+y-1=3.

    Find a normal vector \vec{v} to (2,2,-1) and (1,1,0) for example:

    \vec{v}=(2,2,-1)\times (1,1,0)


    Fernando Revilla
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by questionboy View Post
    3,The point A=(1,2,1,1),B=(2,3,0,3),C=(1,1,2,-1) in R4 from a triangle in R4. find a point D in R4 such ABCD is a parallelogram.

    Find D such that B-A=D-C .


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