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Math Help - 3 true and false questions

  1. #1
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    Question 3 true and false questions

    Question 1
    The plane given by the equation 3x-4y+z-1=0 is perpendicular to the vector
    <9,-12,3> and contains the point (0,0,1).

    The answer is TRUE

    Question 2
    If A is a plane containing the point Psub1, Psub2, and Psub3 is any other point in 3-space, and Psub1Psub2 * Psub1Psub3 =0 , then Psub1Psub3 must be a normal vector for the plane.

    The answer is FALSE

    Question 3
    The four points (-2,1,1), (0,2,3), (1,0,-1), and (5,4,7) all lies in the same plane.

    The answer is TRUE

    Could you please teach me how to solve them? Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    Question 1
    The plane given by the equation 3x-4y+z-1=0 is perpendicular to the vector
    <9,-12,3> and contains the point (0,0,1).

    The answer is TRUE
    ...
    Hello, Jenny,

    your equation is given in normal form. The normal vector which is perpendicular to the palne is: <3, -4, 1>.

    The given vector is 3*<3, -4, 1> = <9, -12, 3> thus this vector is perpendicula to the plane too.

    Plug in the coordinates of the point:
    3*0-4*0+1-1=0 which is true. Thus the point belongs to the plane.

    EB
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  3. #3
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    Quote Originally Posted by Jenny20 View Post
    ...
    Question 3
    The four points (-2,1,1), (0,2,3), (1,0,-1), and (5,4,7) all lies in the same plane.

    The answer is TRUE

    ...
    Hello, Jenny,

    three points, which don't form a straight line or are equal, form a plane:

    Choose one point of the given 4 (I'll take the first one) and two other to set up the equation of this plane:

    <x,y,z>=<-2,1,1>+ r*<<0,2,3>-<-2,1,1>> + s*<<1,0,-1>-<-2,1,1>>, r,s in IR.

    <x,y,z>=<-2,1,1>+ r*<-2,1,2> + s*<3,-1,-2>

    Now plug in the coordinates of the 4th point for <x,y,z>. You'll get 3 equations with 2 variables (r,s). You have to prove that there exist a unique(?) r and s for this system of equations:

    EQU1 : 5 = -2 -2r + 3s
    EQU2 : 4 = 1 + r - s
    EQU3 : 7 = 1 + 2r - 2s

    EQU1 + EQU3 : 12 = -1 +s <==> s = 13. Plug in in EQU3: 7 = 1 + 2r - 2*13
    thus r = 16.

    Now pug in these two values into the unuse EQU2: 4 =1 + 16 - 13
    This is true, therefore the given point belongs to the plane.

    EB
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  4. #4
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    Hello, Jenny!

    Here's another approach to #3 . . .


    3) The four points P(-2,1,1),\:Q(0,2,3),\:R(1,0,-1),\:S(5,4,7) all lie in the same plane.

    Find the plane that contains three of the points, say: P,\:Q,\:R
    . . Then see if S also lies on the plane.

    We have vector \vec{u}\:=\:\overrightarrow{PQ} \:=\:\langle 2,1,2\rangle and vector \vec{v} \:=\:\overrightarrow{PR} \:=\:\langle3,\text{-}1,\text{-}2\rangle

    The normal vector of the plane is perpendicular to both \vec{u} and \vec{v}.

    . . Hence: . \vec{n}\:=\:\begin{vmatrix}i & j & k \\ 2 & 1 & 2 \\ 3 & \text{-}1 & \text{-}2\end{vmatrix} \:=\:\langle 0,10,\text{-}5\rangle \:=\:\langle0,2,\text{-}1\rangle


    The plane through P(-2,1,1) with \vec{n} = \langle0,2,-1\rangle has the equation:

    . . 0(x + 2) + 2(y - 1) - (z - 1) \:=\:0\quad\Rightarrow\quad 2y - z \:=\:1


    Does S(5,4,7) lie on this plane?

    . . 2(4) - 7 \:=\:1 . . . Yes!


    Therefore, the four points are coplanar.

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  5. #5
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    Hello again, Jenny!

    2) If A is a plane containing the points P_1 and P_2
    and P_3 is any other point in 3-space, and: . \overrightarrow{P_1P_2}\cdot\overrightarrow{P_1P_3  } \:=\:0,
    then \vec{P_1P_3} must be a normal vector for the plane.

    The answer is FALSE.

    P_1 and P_2 are on the plane.
    P_3 is any point not on the plane.
    Code:
                        *P3
                        |
                    *---|-------------------*
                   /    |                  /
                  /     |                 /
                 /      * - - - - *      /
                /      P1         P2    /
               / A                     /
              *-----------------------*

    Since \overrightarrow{P_1P_2}\cdot\overrightarrow{P_1P_3  } \:=\:0, then \overrightarrow{P_1P_3} \perp \overrightarrow{P_1P_2}

    But \overrightarrow{P_1P_3} is not necessarily perpendicular to the plane.

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  6. #6
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    Quote Originally Posted by Jenny20 View Post
    ...
    Question 2
    If A is a plane containing the point Psub1, Psub2, and Psub3 is any other point in 3-space, and Psub1Psub2 * Psub1Psub3 =0 , then Psub1Psub3 must be a normal vector for the plane.

    The answer is FALSE

    ...
    Hello, Jenny,

    I assume that the vector product of your problem should be the dot product.

    I've attache a sketch to demonstrate why this statement is false.
    The vector \overrightarrow{Q_1, Q_2} (in black) is perpendicular to the vector \overrightarrow{Q_1, Q_3} (in blue. Q3 doesn't belong to the plan). The normal vector which is perpendicular to the plane (in red) doesn't point to point Q3.

    EB
    Attached Thumbnails Attached Thumbnails 3 true and false questions-jenny_plane1.gif  
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