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Thread: Find parametric equations for the line

  1. #1
    Super Member dhiab's Avatar
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    Find parametric equations for the line

    Find parametric equations for the line through the point (0, 3, 2) that is parallel to the plane x + y + z = 5 and perpendicular to the line x = 1 + t, y = 3 − t, z = 2t. (use the parameter t.)
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    Re: Find parametric equations for the line

    Quote Originally Posted by dhiab View Post
    Find parametric equations for the line through the point (0, 3, 2) that is parallel to the plane x + y + z = 5 and perpendicular to the line x = 1 + t, y = 3 − t, z = 2t. (use the parameter t.)
    The line $<at,3+bt,2+ct>$ contains the point $(0,3,2)$.
    To be parallel to $x+y+z=5$ it must be that $a+b+c=0$
    To be perpendicular to $<1+t,3-t,2t>$ it must intersect the line and $a-b+2c=0$
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    Super Member dhiab's Avatar
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    Re: Find parametric equations for the line

    hello :
    the system : a+b+c=0
    a - b +2c = 0
    3 inconus and 2 equations ?????
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    Re: Find parametric equations for the line

    'inconus'= 'unknowns'

    No, you are misinterpreting what Plato said. Any line in three dimension can be written as x= x_0+ at, y= y_0+ bt, z= z_0+ ct. In order that the line contain the point (0, 3, 2) we must have x= 0+ at= at, y= 3+ bt, and z= 2+ ct. In order that the line be parallel to the plane x+ y+ z= 5, its direction vector <a, b, c> must be perpendicular to the normal vector of the plane, <1, 1, 1>. We must have a+ b+ c= 0. Finally, in order that it be perpendicular to the line x= 1+ t, y= 3- t, z= 2t, it must intersect that line and a- b+ 2c= 0. We have the two equations a+ b+ c= 0 and a- b+ 2c= 0 and the requirement that the two lines must intersect.
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    Re: Find parametric equations for the line

    The point P(0,3,2) is in the plane x+y+z=5 and therefore a line (L) through P and parallel to the plane must be in the plane.

    (L) goes through the point Q(3/2, 5/2, 1) where x = 1 + t, y = 3 − t, z = 2t meets the plane.
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    Super Member dhiab's Avatar
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    Re: Find parametric equations for the line

    Quote Originally Posted by Plato View Post
    The line $<at,3+bt,2+ct>$ contains the point $(0,3,2)$.
    To be parallel to $x+y+z=5$ it must be that $a+b+c=0$
    To be perpendicular to $<1+t,3-t,2t>$ it must intersect the line and $a-b+2c=0$
    hello :
    how to calculate a , b , c ????
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    Re: Find parametric equations for the line

    The line goes through P(0,3,2) and Q(3/2, 5/2, 1)

    with parametric equations

    (x,y,z)=\left(\frac{3}{2}t,3-\frac{1}{2}t,2-t\right)
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    Super Member dhiab's Avatar
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    Re: Find parametric equations for the line

    hello : What do you think of this solution?let A(0,3,2) and (Δ) this line , v vector parallel to (Δ).
    M∈ (Δ) : vector (AM) = t v..... t ∈ R1 )
    (Δ) parallel to the plane x + y + z = 5 : let : n an vector perpendicular to the plane :
    n ⊥ v .... n(1,1,1) so : n.v =0
    means : n.vector (AM) = 0(1)(x)+(1)(y-3)+(1)(z -2) =0 ( vector (AM) = ( x, y -3 , z-2 )x+y+z - 5=0 ...(1)
    2)
    (Δ) perpendicular to the line (Δ') : x = 1+t , y = 3 - t , z = 2t :vector (u) ⊥ v .... vector(u) parallel to (Δ') and vector(u) = (1 , -1 ,1)vector (u) ⊥ vector (AM) means : (1)(x)+(-1)(y-3)+(2)(z -2) =0x - y+2z - 1 = 0 ...(2)so the system : x+y+z - 5=0 ...(1)x - y+2z - 1 = 0 ...(2) (1)+(2) :
    2x+3z - 6 =0x = 3 - (3/2)z subsct in (1) : 3 - (3/2)z +y +z - 5 =0y = 1/2z +2let : z=t an parametric equations for the line (Δ) is : x = 3 - (3/2)t y = (1/2)t +2 z=t
    verifiy : 1) (Δ) parallel to the plane x + y + z = 5 : (-3/2 , 1/2 ,1) perpendicular to (1,1,1)because : (1)(-3/2)+(1)(1/2)+(1)(1) = -1 +1 = 02) (Δ) perpendicular to the line (Δ') : (-3/2 , 1/2 ,1) perpendicular to (1,-1,2)because : (1)(-3/2)+(-1)(1/2)+(1)(2) = -2 +2 = 0A(0, 3, 2)∈(Δ) : 0 = 3-(3/2)t3 = (1/2)t+22 =t same : t = 2
    Last edited by dhiab; Oct 10th 2017 at 09:32 PM.
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